Task C: Eliminate time (\(t\)) to derive the timeless relation \(v^2 = v_0^2 - 2g\Delta y\).
From velocity equation: \(t = \frac{v_0 - v}{g}\). Substitute into displacement:
\(\Delta y = \bar{v}t = \left(\frac{v + v_0}{2}\right)\left(\frac{v_0 - v}{-g}\right) = \frac{v_0^2 - v^2}{-2g} \implies v^2 = v_0^2 - 2g\Delta y\).
Why does a heavy bowling ball fall at the exact same rate of acceleration as a lightweight marble when dropped in a vacuum?
By Newton's 2nd Law, \(a = F_g/m\). Since gravitational force is \(F_g = mg\), \(a = mg/m = g\). The larger force on the bowling ball is perfectly counterbalanced by its larger mass (inertia).
Curriculum Framework: AP Physics 1 Subject: Gravitational Kinematics
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: One-Dimensional Kinematics
Page 2 of 6
TEACHER KEY
Translating algebra into standard AP-style motion maps, spatial coordinate systems, and continuous vector fields.
Strobe interval \(\Delta t = 1.0\text{ s}\). Dots represent position. Vectors show velocity (\(\vec{v}\), blue) and acceleration (\(\vec{a}\), orange).
y = H y = 0 Position Coordinate t = 0 v₀ = 0 t = 1 v₁ a = -g t = 2 v₂ t = 3 v₃
Left track represents the ascent phase. Right track represents the descent phase. Note vector directions!
ASCENT Leg t = 0 t = 1 PEAK (t = 2) v_peak = 0 a = -g DESCENT Leg t = 3 t = 4
Think-Pair-Share Discussion
Discuss with a partner: Why is the acceleration of a projectile at its maximum height exactly \(-9.80\text{ m/s}^2\) and not zero? If acceleration were zero at the peak, how would the object's subsequent motion map look?
1. At what point in Scenario B are the velocity and acceleration vectors parallel? Antiparallel? Explain.
Antiparallel on ascent (\(0 < t < 2\text{ s}\)): \(\vec{v}\) points UP while \(\vec{a}\) points DOWN (slowing down).
Parallel on descent (\(2 < t \le 4\text{ s}\)): both \(\vec{v}\) and \(\vec{a}\) point DOWN (speeding up).
2. If an object is thrown down with an initial velocity \(v_0 = -5 \text{ m/s}\), how does its acceleration vector at \(t=0\) compare to one dropped from rest?
The acceleration vectors are identical: \(\vec{a} = -9.80\text{ m/s}^2\) (downward) for both. Acceleration depends solely on gravity, not the object's initial velocity.
Curriculum Framework: AP Physics 1 Subject: Motion Mapping
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: One-Dimensional Kinematics
Page 3 of 6
TEACHER KEY
A mathematically rigorous analysis of the vertically aligned curves of position, velocity, and acceleration versus time.
Vertically Aligned Curves (Dropped) t y = y₀ 0 y(t) = y₀ - ½gt² t 0 -v v(t) = -gt t 0 a = -g t₁ t₂ t₃
Position Graph Slope: The slope of the tangent line on the position-time (\(y\)-vs-\(t\)) curve represents the instantaneous velocity of the object: \(\text{Slope of } y(t) = v(t)\) As \(t\) increases, the curve curves downward, becoming steeper. This reflects an increasing negative velocity (speeding up in the downward direction).
Velocity Graph Slope: The slope of the velocity-time (\(v\)-vs-\(t\)) line represents the instantaneous acceleration: \(\text{Slope of } v(t) = a = -g\) The constant linear downward slope represents uniform negative gravitational acceleration.
The area bounded between the velocity line and the \(v=0\) axis yields the displacement (\(\Delta y\)) of the falling object.
Because the velocity is negative, this area lies below the time axis and counts as a negative displacement.
Geometrically, the shape is a simple right triangle: \(\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2}(t)(-gt) = -\frac{1}{2}gt^2\)
An object is dropped from a height of \(80 \text{ m}\). Taking \(g = 10 \text{ m/s}^2\): (a) Use the velocity-time area relationship to calculate the displacement of the object after \(4.0\text{ s}\). (b) What is the value of the slope of the velocity-time curve at \(t = 3.5\text{ s}\)? Show your calculations.
(a) At \(t = 4.0\text{ s}\), final velocity is \(v = -gt = -(10)(4) = -40\text{ m/s}\).
\(\text{Area of Triangle} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot (4.0\text{ s}) \cdot (-40\text{ m/s}) = \mathbf{-80\text{ m}}\) (displaced downward).
(b) The slope of any part of a \(v-t\) free-fall graph is constant acceleration. At \(t=3.5\text{ s}\), \(\text{Slope} = a = \mathbf{-10\text{ m/s}^2}\).
Curriculum Framework: AP Physics 1 Subject: Graphical Derivatives
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: One-Dimensional Kinematics
Page 4 of 6
TEACHER KEY
Deconstructing a full projection lifecycle—from launch to peak, and back to release point.
Vertically Aligned Curves (Thrown Upward) t y = 0 y_max t +v₀ 0 -v₀ +Δy -Δy t 0 a = -g t_peak 2 · t_peak
| Phase | Vel / Acc Sign | Physical State |
|---|---|---|
| Ascent | ||
| \(0 < t < t_{peak}\) | \(v > 0\) | |
| \(a = -g\) | Slowing down. Vectors are antiparallel (opposite direction). Net positive displacement. | |
| The Peak | ||
| \(t = t_{peak}\) | \(v = 0\) | |
| \(a = -g\) | Instantaneous rest. Net force is non-zero (gravity still acts). Velocity sign flips. | |
| Descent | ||
| \(t_{peak} < t < 2t_p\) | \(v < 0\) | |
| \(a = -g\) | Speeding up downwards. Vectors are parallel (same direction). Net negative displacement. |
Time Symmetry: The duration of the ascent equals the descent duration: \(t_{peak} = \frac{v_0}{g}\). Total flight time is exactly \(\frac{2v_0}{g}\).
Velocity Symmetry: At any altitude \(y\), the upward velocity \(v_{up}\) and downward velocity \(v_{down}\) have identical magnitudes: \(|v_{up}| = |v_{down}|\).
Geometric Balance: The positive displacement (green triangle) exactly matches the negative displacement (red triangle), confirming a total loop displacement of \(\Delta y = 0\).
Prove analytically, using the timeless equation \(v^2 = v_0^2 - 2g\Delta y\), why the ascending speed at any height \(h\) matches the descending speed at the same height \(h\).
Using the timeless equation: \(v^2 = v_0^2 - 2g\Delta y\), where \(\Delta y = h - y_0 = h\).
Solving for velocity gives: \(v = \pm\sqrt{v_0^2 - 2gh}\).
The positive root corresponds to upward motion (ascent) and the negative root to downward motion (descent). Their magnitudes (speeds) are identical: \(|v_{up}| = |v_{down}| = \sqrt{v_0^2 - 2gh}\).
Curriculum Framework: AP Physics 1 Subject: Symmetrical Motion Analysis
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: One-Dimensional Kinematics
Page 5 of 6
TEACHER KEY
Rigorous, conceptual, and multi-stage physics tasks designed to target common misconceptions in one-dimensional free fall.
PROBLEM 1 The Canyon Drop Challenge (Vertical Displacement)
A physics student drops a heavy stone from rest near the edge of a deep canyon. The stone falls vertically downward through a height of \(H = 80.0 \text{ m}\) before striking the canyon floor. (Define downward as the negative direction, and take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the total time \(t_{fall}\) it takes for the stone to travel from launch to the canyon floor.
(b) If the student instead throws the stone straight downward with an initial speed of \(v_0 = 9.0 \text{ m/s}\), calculate the speed at which the stone strikes the canyon floor.
(a) Calculate the total fall time \(t_{fall}\):
Given: \(y_0 = 80.0\text{ m}\), \(y = 0\), \(v_0 = 0\), \(a = -g = -10.0\text{ m/s}^2\).
Using position equation: \(y = y_0 + v_0 t - \frac{1}{2}gt^2 \implies 0 = 80.0 + 0 - \frac{1}{2}(10.0)t^2\)
\(5.0 t^2 = 80.0 \implies t^2 = 16.0 \implies \mathbf{t_{fall} = 4.0\text{ s}}\).
(b) Determine striking speed with \(v_0 = -9.0\text{ m/s}\):
Using timeless equation: \(v^2 = v_0^2 - 2g\Delta y\) where displacement \(\Delta y = -80.0\text{ m}\):
\(v^2 = (-9.0\text{ m/s})^2 - 2(10.0\text{ m/s}^2)(-80.0\text{ m}) = 81 + 1600 = 1681\).
\(v = -\sqrt{1681} = -41.0\text{ m/s}\). The striking **speed** is the magnitude: \(\mathbf{41.0\text{ m/s}}\).
PROBLEM 2 The Elevated Vertical Toss (Symmetry and Signs)
An object is thrown vertically upward with an initial velocity of \(v_0 = +15.0 \text{ m/s}\) from the edge of a high platform that is exactly \(H = 20.0 \text{ m}\) above the ground. (Take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the maximum height \(y_{max}\) reached by the object relative to ground level.
(b) Determine the velocity vector \(\vec{v}_{impact}\) of the object right before it strikes the ground.
(a) Calculate max height \(y_{max}\):
At peak, \(v = 0\). Find rise height \(\Delta y_{rise}\) above platform: \(v^2 = v_0^2 - 2g\Delta y_{rise} \implies 0^2 = 15.0^2 - 2(10.0)\Delta y_{rise}\).
\(20.0\Delta y_{rise} = 225 \implies \Delta y_{rise} = 11.25\text{ m}\).
Max height relative to ground: \(y_{max} = H + \Delta y_{rise} = 20.0\text{ m} + 11.25\text{ m} = \mathbf{31.25\text{ m}}\).
(b) Determine velocity right before striking the ground:
Using timeless equation directly from launch to impact (\(\Delta y = -20.0\text{ m}\) since ground is 20m below platform):
\(v_{impact}^2 = v_0^2 - 2g\Delta y \implies v_{impact}^2 = 15.0^2 - 2(10.0)(-20.0) = 225 + 400 = 625\).
\(v_{impact} = \pm \sqrt{625} = \pm 25.0\text{ m/s}\). Since the object is falling down: \(\mathbf{\vec{v}_{impact} = -25.0\text{ m/s}}\) (or \(25.0\text{ m/s}\) downward).
Curriculum Framework: AP Physics 1 Subject: Mathematical Rigor Portfolio
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: One-Dimensional Kinematics
Page 6 of 6
TEACHER KEY
Design a standard lab investigation to determine the gravitational acceleration on Earth using linear graphing.
You are tasked with measuring the acceleration due to gravity, \(g\), by dropping an object from rest. From the equipment list below, select the essential tools required for your design, and list the specific quantities they measure.
| Select | Equipment Available | Measured Quantity (Symbol & Unit) |
|---|---|---|
| [ X ] | Meterstick | Drop height (vertical displacement), h, in meters (m) |
| [ X ] | Digital Stopwatch | Fall time duration, t, in seconds (s) |
| [ ] | Electronic Balance | Not needed (mass does not affect free fall acceleration) |
Describe your experimental procedure to determine \(g\). Explain how you will minimize random measurement errors.
1. Measure drop distance h = 0.5 m from the ground using the meterstick.
2. Release a dense sphere (e.g. a steel ball to minimize air resistance) from rest at height h.
3. Use the stopwatch to measure fall time t. Repeat for 5 trials and compute average time t.
4. Repeat steps 1–3 for at least 5 different heights (e.g., 0.8m, 1.1m, 1.4m, 1.7m, 2.0m).
5. To minimize random errors (such as reaction delay), conduct multiple trials at each height and compute averages.
Based on the position-time free fall equation starting from rest, what quantities should be plotted on the vertical axis (\(y\)) and horizontal axis (\(x\)) to obtain a linear graph whose slope can be used to calculate \(g\)? Explain how \(g\) is calculated from that slope.
We start with \(h = \frac{1}{2}gt^2\) (or \(\Delta y = -\frac{1}{2}gt^2\) if negative axis).
We can map this to the linear form \(y = mx + b\):
- **Vertical axis (y):** Plot the drop height \(h\) (in meters).
- **Horizontal axis (x):** Plot the squared time average \(t^2\) (in \(s^2\)).
- **Slope Relationship:** The slope of the resulting linear fit represents \(\text{Slope} = \frac{1}{2}g\).
- **Calculating g:** Rearranging, the acceleration of gravity is calculated as \(\mathbf{g = 2 \cdot \text{slope}}\).
If you used a hand-operated stopwatch to measure time, how would human reaction time affect your calculated value of \(g\)? State whether \(g\) would be an overestimate or an underestimate, and justify your reasoning.
Human reaction delay at impact typically systematically lengthens the measured time (\(t_{measured} > t_{true}\)). Because time is in the denominator of our acceleration relation (\(g = \frac{2h}{t^2}\)), dividing by an artificially large time value will result in a systematic **underestimate** of the value of \(g\) (\(g_{calc} < 9.8\text{ m/s}^2\)).
Curriculum Framework: AP Physics 1 Subject: Experimental Design Questions Key
Task A: Write the equations for horizontal position \(x(t)\) and velocity \(v_x(t)\) given \(a_x = 0\).
Task B: Write the equations for vertical position \(y(t)\) and velocity \(v_y(t)\) given \(a_y = -g\) and \(v_{y0} = 0\).
Task C: Derive the expression for the total flight time \(t_{flight}\) before impact in terms of launch height \(H\) and \(g\).
Think-Pair-Share Discussion
Discuss with a partner: What would happen to the horizontal velocity vector during flight if air resistance were non-negligible? How would the vertical drop time change, if at all?
1. Why do Object A (dropped) and Object B (horizontally launched) hit the floor at the exact same moment if dropped from the same height H?
2. If you double the initial horizontal launch speed \(v_{x0}\) of a projectile, how does this affect (a) flight time, and (b) horizontal range?
Curriculum Framework: AP Physics 1 Subject: Decoupled Trajectory Derivations
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 3 of 6
Name: ______________________ Date: ________
Deconstructing a two-dimensional trajectory into mathematically simple, independent components.
Governed by constant-velocity kinematics because there are no horizontal external forces acting on the projectile.
Acceleration (\(a_x\)): ___________________
Velocity (\(v_x\)): ___________________
Position (\(x\)): ___________________
The horizontal distance covered at any time \(t\) is directly proportional to \(t\). Graphing \(x\) vs \(t\) produces a straight line.
Governed by uniform acceleration kinematics because gravity acts as a constant downward force.
Acceleration (\(a_y\)): ___________________
Velocity (\(v_y\)): ___________________
Position (\(y\)): ___________________
The vertical distance fallen increases quadratically with \(t\). Graphing \(y\) vs \(t\) produces an inverted parabola.
To mathematically prove that the path of a projectile is a parabola, combine your decoupled horizontal position (\(x\)) and vertical position (\(y\)) equations to eliminate time (\(t\)) and solve for \(y(x)\).
Using the horizontal and vertical position equations, derive an expression for the total horizontal range \(R\) in terms of launch velocity \(v_{x0}\), height \(H\), and gravity \(g\). Explain how range is affected by doubling both height \(H\) and launch speed \(v_{x0}\).
Curriculum Framework: AP Physics 1 Subject: Trajectory Derivations
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 4 of 6
Name: ______________________ Date: ________
Active practice sketching key horizontal and vertical kinematic curves over flight time.
Horizontal Graphs (x-direction) t x t v_x v_x₀ t a_x t_flight
Vertical Graphs (y-direction) t y H t v_y 0 t a_y -g t_flight
Review the horizontal and vertical velocity graphs. How does the area under the \(v_x(t)\) graph relate to the total horizontal range, and how does the area under the \(v_y(t)\) graph relate to the launch height \(H\)? Draw geometric shapes on your paper to verify.
Curriculum Framework: AP Physics 1 Subject: Graphical Decoupling Practice
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 5 of 6
Name: ______________________ Date: ________
Medium-hard numerical problems standard in AP Physics 1 curriculum testing component mechanics.
PROBLEM 1 The Air drop Target Intercept
A supply plane flying horizontally with a constant speed of \(v_{x0} = 40.0 \text{ m/s}\) at an altitude of \(H = 45.0 \text{ m}\) releases a heavy rescue package. The package falls freely until landing near a stranded climber on the ground. (Take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the total time \(t\) that the package is in the air.
(b) Determine the horizontal distance \(d\) from the climber that the plane must release the package.
(c) Determine the total impact speed of the package immediately prior to hitting the ground.
PROBLEM 2 The Lab Table Roll-Off
A steel marble rolls off a horizontal lab bench of height \(1.25 \text{ m}\) and lands on the floor a horizontal range of exactly \(R = 1.80 \text{ m}\) from the table edge. (Take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the total time the marble is in flight.
(b) Calculate the initial horizontal velocity \(v_{x0}\) as it rolled off the table.
(c) Determine the magnitude of the velocity vector right before impact with the floor.
Curriculum Framework: AP Physics 1 Subject: Component Problem Solving
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 6 of 6
Name: ______________________ Date: ________
Design a standard lab investigation to determine the relationship between launch velocity and range.
You are tasked with measuring how the initial horizontal velocity \(v_{x0}\) affects the horizontal range \(R\) of a ball rolled off a table of constant height \(H\). Select your essential tools below and list the quantities they measure.
| Select | Equipment Available | Measured Quantity (Symbol & Unit) |
|---|---|---|
| Meterstick | ____________________________________________________ | |
| Photogate System | ____________________________________________________ | |
| Carbon Landing Paper | ____________________________________________________ |
Describe your experimental procedure to systematically vary launch velocity and measure horizontal range.
Using the range expression you derived on Page 3, describe what you would plot on the vertical and horizontal axes to obtain a linear graph whose slope could be used to calculate gravity, \(g\).
State one systematic source of error in this experiment that would cause your range measurements to be systematically smaller than predicted. How does this affect your calculated value of \(g\)?
Curriculum Framework: AP Physics 1 Subject: Experimental Design Questions
Using the coordinate definitions (upward and rightward as positive) and initial launch parameters, derive the algebraic kinematics equations.
Task A: Write the equations for horizontal position \(x(t)\) and velocity \(v_x(t)\) given \(a_x = 0\).
Since horizontal acceleration is zero (\(a_x = 0\)):
Velocity is constant: \(v_x(t) = v_{x0}\). Position is linear: \(x(t) = x_0 + v_{x0}t\) (assuming \(x_0 = 0 \implies x(t) = v_{x0}t\)).
Task B: Write the equations for vertical position \(y(t)\) and velocity \(v_y(t)\) given \(a_y = -g\) and \(v_{y0} = 0\).
Since vertical acceleration is constant downward (\(a_y = -g\)) and initial vertical velocity is zero (\(v_{y0} = 0\)):\ or falling from rest:
Velocity is linear: \(v_y(t) = -gt\). Position is quadratic: \(y(t) = y_0 + v_{y0}t - \frac{1}{2}gt^2 \implies y(t) = H - \frac{1}{2}gt^2\).
Task C: Derive the expression for the total flight time \(t_{flight}\) before impact in terms of launch height \(H\) and \(g\).
Impact occurs when vertical position \(y(t) = 0\). Set equation equal to zero:
\(0 = H - \frac{1}{2}gt_{flight}^2 \implies \frac{1}{2}gt_{flight}^2 = H \implies t_{flight}^2 = \frac{2H}{g} \implies \mathbf{t_{flight} = \sqrt{\frac{2H}{g}}}\).
Think-Pair-Share Discussion
Discuss with a partner: What would happen to the horizontal velocity vector during flight if air resistance were non-negligible? How would the vertical drop time change, if at all?
1. Why do Object A (dropped) and Object B (horizontally launched) hit the floor at the exact same moment if dropped from the same height H?
Both start with zero initial vertical velocity (\(v_{y0}=0\)) and experience the same downward vertical acceleration \(a_y = -g\). Since their vertical motion is identical, they fall the same height \(H\) in the exact same time.
2. If you double the initial horizontal launch speed \(v_{x0}\) of a projectile, how does this affect (a) flight time, and (b) horizontal range?
(a) Flight time remains unchanged, as \(t_{flight} = \sqrt{2H/g}\) is independent of horizontal speed.
(b) Since range \(R = v_{x0}t_{flight}\), doubling \(v_{x0}\) while time is constant exactly doubles horizontal range.
Curriculum Framework: AP Physics 1 Subject: Decoupled Trajectory Derivations
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 3 of 6
TEACHER KEY
Deconstructing a two-dimensional trajectory into mathematically simple, independent components.
Governed by constant-velocity kinematics because there are no horizontal external forces acting on the projectile.
Acceleration (\(a_x\)): 0
Velocity (\(v_x\)): v_x₀ (constant)
Position (\(x\)): x = v_x₀ · t
The horizontal distance covered at any time \(t\) is directly proportional to \(t\). Graphing \(x\) vs \(t\) produces a straight line.
Governed by uniform acceleration kinematics because gravity acts as a constant downward force.
Acceleration (\(a_y\)): -g ≈ -9.80 m/s²
Velocity (\(v_y\)): v_y = -gt
Position (\(y\)): y = H - ½gt²
The vertical distance fallen increases quadratically with \(t\). Graphing \(y\) vs \(t\) produces an inverted parabola.
To mathematically prove that the path of a projectile is a parabola, combine your decoupled horizontal position (\(x\)) and vertical position (\(y\)) equations to eliminate time (\(t\)) and solve for \(y(x)\).
1. From horizontal position: \(x = v_{x0}t \implies t = \frac{x}{v_{x0}}\).
2. Substitute this expression for \(t\) into vertical position equation \(y = H - \frac{1}{2}gt^2\):
\(y(x) = H - \frac{1}{2}g\left(\frac{x}{v_{x0}}\right)^2 \implies \mathbf{y(x) = H - \left(\frac{g}{2v_{x0}^2}\right)x^2}\)
3. Since \(H\), \(g\), and \(v_{x0}\) are constants, this matches the standard quadratic form of a downward opening parabola \(y = C_1 - C_2 x^2\) with vertex at \((0, H)\).
Using the horizontal and vertical position equations, derive an expression for the total horizontal range \(R\) in terms of launch velocity \(v_{x0}\), height \(H\), and gravity \(g\). Explain how range is affected by doubling both height \(H\) and launch speed \(v_{x0}\).
Substitute \(t = \sqrt{2H/g}\) into \(x = v_{x0}t\) to get horizontal range: \(\mathbf{R = v_{x0}\sqrt{\frac{2H}{g}}}\).
- Doubling launch speed \(v_{x0}\) increases range by a factor of 2.
- Doubling height \(H\) increases range by a factor of \(\sqrt{2} \approx 1.41\).
- Doubling both increases range by a factor of \(2\sqrt{2} \approx \mathbf{2.83}\) times.
Curriculum Framework: AP Physics 1 Subject: Trajectory Derivations
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 4 of 6
TEACHER KEY
The official key curves drawn in red representing position, velocity, and acceleration versus time.
Horizontal Graphs (x-direction) t x x = v_x₀ · t t v_x v_x₀ t a_x t_flight
Vertical Graphs (y-direction) t y H y = H - ½gt² t v_y 0 v_y = -gt t a_y -g t_flight
Review the horizontal and vertical velocity graphs. How does the area under the \(v_x(t)\) graph relate to the total horizontal range, and how does the area under the \(v_y(t)\) graph relate to the launch height \(H\)? Draw geometric shapes on your paper to verify.
- The area under the \(v_x(t)\) curve forms a rectangle: \(\text{Area} = v_{x0} \cdot t_{flight} = \mathbf{\text{horizontal range (R)}}\).
- The area under the \(v_y(t)\) curve forms a triangle: \(\text{Area} = \frac{1}{2}(t_{flight})(-gt_{flight}) = -\frac{1}{2}gt_{flight}^2 = \mathbf{-H}\) (downward displacement equal to launch height).
Curriculum Framework: AP Physics 1 Subject: Graphical Decoupling Practice Key
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 5 of 6
TEACHER KEY
Medium-hard numerical problems standard in AP Physics 1 curriculum testing component mechanics.
PROBLEM 1 The Air drop Target Intercept
A supply plane flying horizontally with a constant speed of \(v_{x0} = 40.0 \text{ m/s}\) at an altitude of \(H = 45.0 \text{ m}\) releases a heavy rescue package. (Take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the total time \(t\) that the package is in the air.
(b) Determine the horizontal distance \(d\) from the climber that the plane must release the package.
(c) Determine the total impact speed of the package immediately prior to hitting the ground.
(a) Calculate total flight time \(t\):
Using vertical position with \(y_0 = H = 45.0\text{ m}\), \(v_{y0}=0\), and \(y_{final}=0\):
\(y = y_0 + v_{y0}t - \frac{1}{2}gt^2 \implies 0 = 45.0 + 0 - 5.0t^2 \implies 5.0t^2 = 45.0 \implies t^2 = 9.0 \implies \mathbf{t = 3.0\text{ s}}\).
(b) Determine horizontal release distance \(d\):
Using horizontal position with constant velocity \(v_x = v_{x0} = 40.0\text{ m/s}\):
\(d = v_{x0}t = (40.0\text{ m/s})(3.0\text{ s}) = \mathbf{120.0\text{ m}}\).
(c) Determine resultant impact speed \(v\):
Horizontal velocity component remains constant: \(v_x = 40.0\text{ m/s}\).
Vertical velocity component right before impact: \(v_y = v_{y0} - gt = 0 - (10.0\text{ m/s}^2)(3.0\text{ s}) = -30.0\text{ m/s}\).
Impact speed: \(v = \sqrt{v_x^2 + v_y^2} = \sqrt{40.0^2 + (-30.0)^2} = \sqrt{2500} = \mathbf{50.0\text{ m/s}}\).
PROBLEM 2 The Lab Table Roll-Off
A steel marble rolls off a horizontal lab bench of height \(1.25 \text{ m}\) and lands on the floor a horizontal range of exactly \(R = 1.80 \text{ m}\) from the table edge. (Take \(g = 10.0 \text{ m/s}^2\)).
(a) Calculate the total time the marble is in flight.
(b) Calculate the initial horizontal velocity \(v_{x0}\) as it rolled off the table.
(c) Determine the magnitude of the velocity vector right before impact with the floor.
(a) Calculate total flight time \(t\):
\(t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2(1.25\text{ m})}{10.0\text{ m/s}^2}} = \sqrt{0.25\text{ s}^2} = \mathbf{0.50\text{ s}}\).
(b) Determine initial horizontal velocity \(v_{x0}\):
\(R = v_{x0}t \implies 1.80\text{ m} = v_{x0}(0.50\text{ s}) \implies v_{x0} = \frac{1.80}{0.50} = \mathbf{3.60\text{ m/s}}\).
(c) Determine impact velocity magnitude \(v\):
\(v_x = 3.60\text{ m/s}\). Vertical velocity component: \(v_y = -gt = -(10.0)(0.50) = -5.00\text{ m/s}\).
\(v = \sqrt{3.60^2 + (-5.00)^2} = \sqrt{12.96 + 25.00} = \sqrt{37.96} \approx \mathbf{6.16\text{ m/s}}\).
Curriculum Framework: AP Physics 1 Subject: Component Problem Solving
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 6 of 6
TEACHER KEY
Design a standard lab investigation to determine the relationship between launch velocity and range.
You are tasked with measuring how the initial horizontal velocity \(v_{x0}\) affects the horizontal range \(R\) of a ball rolled off a table of constant height \(H\). Select your essential tools below and list the quantities they measure.
| Select | Equipment Available | Measured Quantity (Symbol & Unit) |
|---|---|---|
| [ X ] | Meterstick | Table height H and horizontal drop range R (in meters) |
| [ X ] | Photogate System | Initial horizontal speed v_x₀ right as it exits the ramp (m/s) |
| [ X ] | Carbon Landing Paper | Marks the exact impact point on the floor to measure range R |
Describe your experimental procedure to systematically vary launch velocity and measure horizontal range.
1. Set up a launch ramp on a horizontal lab table of height H. Measure height H with a meterstick.
2. Place a photogate at the very end of the ramp (horizontal section) to record exit speed v_x₀.
3. Place carbon paper over plain paper on the floor where the ball is expected to land.
4. Release the sphere from a designated height on the ramp, record the v_x₀ exit speed, and measure the horizontal range R from table edge to landing mark.
5. Repeat for 5 different release heights, performing 3 trials at each height to compute average range R.
Using the equation \(R = v_{x0} \sqrt{\frac{2H}{g}}\), describe what you would plot on the vertical and horizontal axes to obtain a linear graph whose slope could be used to calculate gravity, \(g\).
We map \(R = \left(\sqrt{\frac{2H}{g}}\right) v_{x0}\) to the linear form \(y = mx + b\):
- **Vertical axis (y):** Plot the horizontal range \(R\) (in meters).
- **Horizontal axis (x):** Plot the initial horizontal speed \(v_{x0}\) (in m/s).
- **Slope Relationship:** The slope of the resulting linear graph is equal to: \(\text{Slope} = \sqrt{\frac{2H}{g}}\).
- **Calculating g:** Square both sides: \(\text{Slope}^2 = \frac{2H}{g} \implies \mathbf{g = \frac{2H}{\text{Slope}^2}}\).
State one systematic source of error in this experiment that would cause your range measurements to be systematically smaller than predicted. How does this affect your calculated value of \(g\)?
Frictional forces acting as the ball rolls off the table edge or air resistance during flight would systematically reduce the actual horizontal range (\(R_{measured} < R_{predicted}\)). Since the manipulated range \(R\) is too small, the slope of the \(R\) vs. \(v_{x0}\) graph is artificially low. Since \(g = \frac{2H}{\text{Slope}^2}\), an artificially small slope value will result in a systematic **overestimate** of \(g\) (\(g_{calc} > 9.8\text{ m/s}^2\)).
Curriculum Framework: AP Physics 1 Subject: Experimental Design Questions Key
Using upward and rightward as positive coordinate directions, write out the explicit expressions for position and velocity.
Task A: Write the general equations for horizontal position \(x(t)\) and velocity \(v_x(t)\) in terms of launch parameters \(v_0\) and \(\theta\).
Task B: Write the general equations for vertical position \(y(t)\) and velocity \(v_y(t)\) in terms of launch parameters \(v_0\), \(\theta\), and acceleration \(g\).
Task C: Derive the expression for the time-to-peak \(t_{peak}\) using the condition \(v_y(t_{peak}) = 0\). Then write the expression for total symmetrical flight time \(t_{flight}\).
Think-Pair-Share Discussion
Discuss with a partner: At the peak of an angled trajectory, the vertical velocity of the projectile is zero, but the vertical acceleration is \(a_y = -g \neq 0\). Explain in terms of forces and vectors why the acceleration is not zero, and describe what the projectile's path would look like if the acceleration suddenly became zero at the peak.
1. Max Components At what angle \(\theta\) is horizontal velocity maximized? At what angle is vertical velocity maximized? How does this dictate peak height?
2. Complementary Symmetry For a level flat launch, explain why launching at angle \(\theta\) and launch angle \(90^\circ - \theta\) results in the exact same horizontal range.
Curriculum Framework: AP Physics 1 Subject: Decoupled Equations Derivations
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 3 of 6
Name: ______________________ Date: ________
Deconstructing an angled trajectory mathematically to reveal its parabolic path and horizontal reach limits.
Governed by constant-velocity kinematics because there are no horizontal external forces acting on the projectile.
Acceleration (\(a_x\)): ___________________
Velocity (\(v_x\)): ___________________
Position (\(x\)): ___________________
The horizontal position covered at any time \(t\) is linear. The projectile moves outward at a fixed speed \(v_0 \cos\theta\).
Governed by uniform acceleration kinematics because gravity acts as a constant downward force.
Acceleration (\(a_y\)): ___________________
Velocity (\(v_y\)): ___________________
Position (\(y\)): ___________________
The vertical height is quadratic in time. The trajectory starts with upward speed, halts, and drops down.
Using the condition that at the peak of flight, the vertical velocity of the projectile is zero (\(v_y(t_{peak}) = 0\)), use your vertical velocity and position equations to derive an expression for the maximum peak height \(H_{max}\) in terms of initial velocity \(v_0\), launch angle \(\theta\), and gravity \(g\).
Using the symmetrical total flight time \(t_{flight} = \frac{2 v_0 \sin\theta}{g}\) derived on Page 2, substitute this expression into the horizontal position equation \(x(t)\). Apply the trigonometric identity \(2\sin\theta\cos\theta = \sin(2\theta)\) to derive the flat-ground horizontal Range equation \(R(\theta)\). Explain mathematically why the maximum possible range occurs at \(\theta = 45^\circ\).
Curriculum Framework: AP Physics 1 Subject: Trajectory & Symmetrical Range Equations
AP Physics 1 Unit 1: Two-Dimensional Kinematics
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Name: ______________________ Date: ________
Active practice sketching key horizontal and vertical kinematic curves over flat-field flight time.
Horizontal Graphs (x-direction) t x t v_x v_0 cos θ t a_x t_peak t_flight
Vertical Graphs (y-direction) t y H_max t v_y v_0 sin θ -v_0 sin θ t a_y -g t_peak t_flight
Observe the slope of the vertical velocity graph, \(v_y(t)\). Explain why the slope is constant throughout the entire flight, including at the exact instant \(t = t_{peak}\) when \(v_y = 0\). Relate this slope value to the vertical acceleration.
Curriculum Framework: AP Physics 1 Subject: Graphical Angled Motion Practice
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 5 of 6
Name: ______________________ Date: ________
Medium-hard component kinematics problems standard in the AP Physics 1 curriculum.
PROBLEM 1 The Symmetrical Soccer Kick
A soccer ball is kicked from level ground with an initial speed of \(v_0 = 20.0 \text{ m/s}\) at an angle of \(\theta = 37.0^\circ\) above the horizontal. (Take \(g = 10.0 \text{ m/s}^2\), \(\cos(37^\circ) = 0.80\), and \(\sin(37^\circ) = 0.60\)).
(a) Decompose the initial velocity vector to find the horizontal component \(v_{x0}\) and vertical component \(v_{y0}\).
(b) Calculate the maximum height \(H_{max}\) reached by the soccer ball above the ground.
(c) Determine the total flight time \(t_{flight}\) and the total horizontal range \(R\) covered before landing.
PROBLEM 2 The Elevated Cliff-Edge Launch
A small steel projectile is launched from the edge of an elevated cliff of height \(H = 40.0 \text{ m}\) above a flat valley. The initial launch velocity is \(v_0 = 25.0 \text{ m/s}\) directed at an angle of \(\theta = 30.0^\circ\) above the horizontal. (Take \(g = 10.0 \text{ m/s}^2\), \(\cos(30^\circ) = 0.87\), and \(\sin(30^\circ) = 0.50\)).
(a) Find the initial horizontal and vertical velocity components.
(b) Calculate the total flight time \(t_{flight}\) before the projectile strikes the valley ground.
(c) Determine the total horizontal displacement from the base of the cliff.
Curriculum Framework: AP Physics 1 Subject: Symmetrical & Elevated Problem Solving
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 6 of 6
Name: ______________________ Date: ________
Design an experimental investigation to study the relationship between launch angle and horizontal range.
You are tasked with measuring how the initial launch angle \(\theta\) affects the horizontal range \(R\) of a projectile launched with a constant speed \(v_0\) from flat ground. Select your essential tools below and list the quantities they measure.
| Select | Equipment Available | Measured Quantity (Symbol & Unit) |
|---|---|---|
| Meterstick | ____________________________________________________ | |
| Spring Launcher with Protractor | ____________________________________________________ | |
| Carbon Impact Landing Paper | ____________________________________________________ |
Describe your experimental procedure to systematically vary launch angle and measure horizontal range.
Using the symmetrical range equation \(R = \frac{v_0^2}{g} \sin(2\theta)\), describe what variables you should plot on the vertical and horizontal axes to obtain a linear graph whose slope could be used to calculate the initial launch speed, \(v_0\).
Under real atmospheric conditions, air resistance operates as a non-negligible force. Would you expect the angle that maximizes horizontal range to be slightly greater than, equal to, or slightly less than \(45^\circ\)? Justify your physical reasoning.
Curriculum Framework: AP Physics 1 Subject: Experimental Design Questions
Using upward and rightward as positive coordinate directions, write out the explicit expressions for position and velocity.
Task A: Write the general equations for horizontal position \(x(t)\) and velocity \(v_x(t)\) in terms of launch parameters \(v_0\) and \(\theta\).
Since horizontal acceleration is zero (\(a_x = 0\)):
Velocity is constant: \(\mathbf{v_x(t) = v_{x0} = v_0 \cos\theta}\).
Position is linear: \(\mathbf{x(t) = v_{x0}t = v_0 \cos\theta \cdot t}\).
Task B: Write the general equations for vertical position \(y(t)\) and velocity \(v_y(t)\) in terms of launch parameters \(v_0\), \(\theta\), and acceleration \(g\).
Since vertical acceleration is constant downward (\(a_y = -g\)):
Velocity is linear: \(\mathbf{v_y(t) = v_0 \sin\theta - gt}\).
Position is quadratic: \(\mathbf{y(t) = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2}\) (assuming ground-to-ground \(y_0 = 0\)).
Task C: Derive the expression for the time-to-peak \(t_{peak}\) using the condition \(v_y(t_{peak}) = 0\). Then write the expression for total symmetrical flight time \(t_{flight}\).
Set \(v_y(t_{peak}) = 0 \implies v_0 \sin\theta - g t_{peak} = 0 \implies \mathbf{t_{peak} = \frac{v_0 \sin\theta}{g}}\).
Due to symmetrical ground-to-ground flight, the total flight time is exactly twice the rise time:
\(\mathbf{t_{flight} = 2 t_{peak} = \frac{2 v_0 \sin\theta}{g}}\).
Think-Pair-Share Discussion
Discuss with a partner: At the peak of an angled trajectory, the vertical velocity of the projectile is zero, but the vertical acceleration is \(a_y = -g \neq 0\). Explain in terms of forces and vectors why the acceleration is not zero, and describe what the projectile's path would look like if the acceleration suddenly became zero at the peak.
1. Max Components At what angle \(\theta\) is horizontal velocity maximized? At what angle is vertical velocity maximized? How does this dictate peak height?
Horizontal velocity (\(v_0 \cos\theta\)) is maximized at \(\mathbf{\theta = 0^\circ}\) (\(\cos 0^\circ=1\)). Vertical velocity (\(v_0 \sin\theta\)) is maximized at \(\mathbf{\theta = 90^\circ}\) (\(\sin 90^\circ=1\)). Since peak height depends only on vertical velocity, launching straight up (\(90^\circ\)) yields maximum peak height.
2. Complementary Symmetry For a level flat launch, explain why launching at angle \(\theta\) and launch angle \(90^\circ - \theta\) results in the exact same horizontal range.
Range depends on \(\sin(2\theta)\). Substituting \(90^\circ-\theta\) gives \(\sin(2(90^\circ-\theta)) = \sin(180^\circ-2\theta)\). Since \(\sin(180^\circ-\alpha)=\sin\alpha\), the factor \(\sin(2\theta)\) remains identical, producing equal horizontal ranges!
Curriculum Framework: AP Physics 1 Subject: Decoupled Equations Derivations Key
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 3 of 6
TEACHER KEY
Deconstructing an angled trajectory mathematically to reveal its parabolic path and horizontal reach limits.
Governed by constant-velocity kinematics because there are no horizontal external forces acting on the projectile.
Acceleration (\(a_x\)): 0
Velocity (\(v_x\)): v_₀ cos θ (constant)
Position (\(x\)): x = v_₀ cos θ · t
The horizontal position covered at any time \(t\) is linear. The projectile moves outward at a fixed speed \(v_0 \cos\theta\).
Governed by uniform acceleration kinematics because gravity acts as a constant downward force.
Acceleration (\(a_y\)): -g ≈ -9.80 m/s²
Velocity (\(v_y\)): v_y = v_₀ sin θ - gt
Position (\(y\)): y = v_₀ sin θ · t - ½gt²
The vertical height is quadratic in time. The trajectory starts with upward speed, halts, and drops down.
Using the condition that at the peak of flight, the vertical velocity of the projectile is zero (\(v_y(t_{peak}) = 0\)), use your vertical velocity and position equations to derive an expression for the maximum peak height \(H_{max}\) in terms of initial velocity \(v_0\), launch angle \(\theta\), and gravity \(g\).
1. Using vertical velocity: \(v_y(t) = v_0 \sin\theta - gt \implies \text{At peak: } t_{peak} = \frac{v_0 \sin\theta}{g}\).
2. Substitute \(t_{peak}\) into vertical position: \(H_{max} = y(t_{peak}) = v_0 \sin\theta \left(\frac{v_0 \sin\theta}{g}\right) - \frac{1}{2}g \left(\frac{v_0 \sin\theta}{g}\right)^2\).
\(H_{max} = \frac{v_0^2 \sin^2\theta}{g} - \frac{v_0^2 \sin^2\theta}{2g} \implies \mathbf{H_{max} = \frac{v_0^2 \sin^2\theta}{2g}}\)
Alternative Method: Use \(v_y^2 = v_{y0}^2 - 2g \Delta y\). At peak, \(0 = (v_0\sin\theta)^2 - 2g H_{max} \implies \mathbf{H_{max} = \frac{v_0^2 \sin^2\theta}{2g}}\).
Using the symmetrical total flight time \(t_{flight} = \frac{2 v_0 \sin\theta}{g}\) derived on Page 2, substitute this expression into the horizontal position equation \(x(t)\). Apply the trigonometric identity \(2\sin\theta\cos\theta = \sin(2\theta)\) to derive the flat-ground horizontal Range equation \(R(\theta)\). Explain mathematically why the maximum possible range occurs at \(\theta = 45^\circ\).
\(R = x(t_{flight}) = v_0 \cos\theta \cdot \left(\frac{2 v_0 \sin\theta}{g}\right) = \frac{v_0^2}{g} \cdot (2\sin\theta\cos\theta)\).
Applying double-angle identity: \(\mathbf{R = \frac{v_0^2 \sin(2\theta)}{g}}\).
The function \(R(\theta)\) is maximized when the sine term \(\sin(2\theta) = 1\) (its peak value).
This occurs when its argument: \(2\theta = 90^\circ \implies \mathbf{\theta = 45^\circ}\). Any launch angle above or below \(45^\circ\) results in a smaller horizontal reach.
Curriculum Framework: AP Physics 1 Subject: Trajectory & Symmetrical Range Equations Key
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 4 of 6
TEACHER KEY
Active practice sketching key horizontal and vertical kinematic curves over flat-field flight time.
Horizontal Graphs (x-direction) t x Slope = v_₀ cos θ t v_x v_0 cos θ t a_x t_peak t_flight
Vertical Graphs (y-direction) t y H_max t v_y v_0 sin θ -v_0 sin θ Slope = -g t a_y -g t_peak t_flight
Observe the slope of the vertical velocity graph, \(v_y(t)\). Explain why the slope is constant throughout the entire flight, including at the exact instant \(t = t_{peak}\) when \(v_y = 0\). Relate this slope value to the vertical acceleration.
The vertical velocity graph is a straight descending line with constant slope of \(\mathbf{-g \approx -10.0\text{ m/s}^2}\). The slope is constant because gravity is a continuous uniform force exerting downward acceleration. Even at \(t = t_{peak}\), the object's state of motion is changing, meaning its velocity is transitioning from positive to negative at exactly the rate of \(-g\) units per second.
Curriculum Framework: AP Physics 1 Subject: Graphical Angled Motion Practice Key
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 5 of 6
TEACHER KEY
Medium-hard component kinematics problems standard in the AP Physics 1 curriculum.
PROBLEM 1 The Symmetrical Soccer Kick
A soccer ball is kicked from level ground with an initial speed of \(v_0 = 20.0 \text{ m/s}\) at an angle of \(\theta = 37.0^\circ\) above the horizontal. (Take \(g = 10.0 \text{ m/s}^2\), \(\cos(37^\circ) = 0.80\), and \(\sin(37^\circ) = 0.60\)).
(a) Decompose the initial velocity vector to find the horizontal component \(v_{x0}\) and vertical component \(v_{y0}\).
(b) Calculate the maximum height \(H_{max}\) reached by the soccer ball above the ground.
(c) Determine the total flight time \(t_{flight}\) and the total horizontal range \(R\) covered before landing.
(a) Find velocity components:
\(v_{x0} = v_0 \cos(37^\circ) = (20.0\text{ m/s})(0.80) = \mathbf{16.0\text{ m/s}}\)
\(v_{y0} = v_0 \sin(37^\circ) = (20.0\text{ m/s})(0.60) = \mathbf{12.0\text{ m/s}}\)
(b) Calculate maximum peak height \(H_{max}\):
Use third kinematic equation vertically: \(v_y^2 = v_{y0}^2 - 2g \Delta y\). At peak, \(v_y = 0\):
\(0 = (12.0)^2 - 2(10.0)H_{max} \implies 20.0 H_{max} = 144 \implies \mathbf{H_{max} = 7.20\text{ m}}\).
(c) Find flight time and horizontal range:
\(t_{flight} = \frac{2v_{y0}}{g} = \frac{2(12.0\text{ m/s})}{10.0\text{ m/s}^2} = \mathbf{2.40\text{ s}}\).
\(R = v_{x0} \cdot t_{flight} = (16.0\text{ m/s})(2.40\text{ s}) = \mathbf{38.4\text{ m}}\).
(Verification via Range Equation: \(R = \frac{20.0^2 \sin(74^\circ)}{10.0} = 40.0 \cdot 2(0.8)(0.6) = 38.4\text{ m}\). Perfect check!)
PROBLEM 2 The Elevated Cliff-Edge Launch
A small steel projectile is launched from the edge of an elevated cliff of height \(H = 40.0 \text{ m}\) above a flat valley. The initial launch velocity is \(v_0 = 25.0 \text{ m/s}\) directed at an angle of \(\theta = 30.0^\circ\) above the horizontal. (Take \(g = 10.0 \text{ m/s}^2\), \(\cos(30^\circ) = 0.87\), and \(\sin(30^\circ) = 0.50\)).
(a) Find the initial horizontal and vertical velocity components.
(b) Calculate the total flight time \(t_{flight}\) before the projectile strikes the valley ground.
(c) Determine the total horizontal displacement from the base of the cliff.
(a) Decompose: \(v_{x0} = 25.0 \cos(30^\circ) = \mathbf{21.75\text{ m/s}}\) and \(v_{y0} = 25.0 \sin(30^\circ) = \mathbf{12.5\text{ m/s}}\).
(b) Flight time: \(\Delta y = -40.0\text{ m}\) (displacement is negative since landing below cliff).
\(\Delta y = v_{y0} t - \frac{1}{2}gt^2 \implies -40.0 = 12.5t - 5.0t^2 \implies 5.0t^2 - 12.5t - 40.0 = 0 \implies t^2 - 2.5t - 8.0 = 0\).
Solving quadratic: \(t = \frac{2.5 \pm \sqrt{(-2.5)^2 - 4(1)(-8.0)}}{2} = \frac{2.5 \pm \sqrt{38.25}}{2} \approx \frac{2.5 \pm 6.18}{2} \implies \mathbf{t_{flight} \approx 4.34\text{ s}}\).
(c) Horizontal range: \(R = v_{x0} \cdot t_{flight} = (21.75\text{ m/s})(4.34\text{ s}) = \mathbf{94.4\text{ m}}\).
Curriculum Framework: AP Physics 1 Subject: Symmetrical & Elevated Problem Solving Key
Official Teacher Answer Key — Reference Only
AP Physics 1 Unit 1: Two-Dimensional Kinematics
Page 6 of 6
TEACHER KEY
Design an experimental investigation to study the relationship between launch angle and horizontal range.
You are tasked with measuring how the initial launch angle \(\theta\) affects the horizontal range \(R\) of a projectile launched with a constant speed \(v_0\) from flat ground. Select your essential tools below and list the quantities they measure.
| Select | Equipment Available | Measured Quantity (Symbol & Unit) |
|---|---|---|
| [ X ] | Meterstick | The horizontal drop distance (Range, R) from launcher muzzle to mark (m) |
| [ X ] | Spring Launcher with Protractor | Alters the launch angle theta (degrees) while holding muzzle velocity v_₀ constant |
| [ X ] | Carbon Impact Landing Paper | Marks the exact, repeatable landing spot on the flat floor plane for range calculations |
Describe your experimental procedure to systematically vary launch angle and measure horizontal range.
1. Anchor the spring launcher to a flat floor. Adjust the angle \(\theta = 15.0^\circ\) with the launcher's built-in protractor.
2. Set the spring level to a specific tension setting and maintain this constant launch setting throughout the lab.
3. Secure plain paper and carbon paper on the landing floor area at the predicted landing distance.
4. Fire the projectile. Locate the carbon impact dot and measure the straight distance \(R\) from launcher exit to the dot.
5. Repeat for 3 trials to obtain an average range. Increase the angle by \(10^\circ\) increments up to \(75^\circ\) and repeat steps.
Using the symmetrical range equation \(R = \frac{v_0^2}{g} \sin(2\theta)\), describe what variables you should plot on the vertical and horizontal axes to obtain a linear graph whose slope could be used to calculate the initial launch speed, \(v_0\).
Represent range as a linear equation: \(y = mx + b \implies R = \left(\frac{v_0^2}{g}\right) \sin(2\theta)\):
- **Vertical y-axis:** Plot the average horizontal range \(R\) (meters).
- **Horizontal x-axis:** Plot the calculated trigonometric term \(\sin(2\theta)\) (dimensionless).
- **Slope of linear curve:** Plotting \(R\) versus \(\sin(2\theta)\) produces a straight line where \(\text{Slope} = \frac{v_0^2}{g}\).
- **Solving for initial speed:** Multiply slope by \(g\) and take the square root: \(\mathbf{v_0 = \sqrt{\text{Slope} \cdot g}}\).
Under real atmospheric conditions, air resistance operates as a non-negligible force. Would you expect the angle that maximizes horizontal range to be slightly greater than, equal to, or slightly less than \(45^\circ\)? Justify your physical reasoning.
Air resistance opposes motion, slowing the horizontal velocity over time. To maximize range in real atmosphere, we must launch slightly flatter (value **less than \(45^\circ\)**, typically around \(38^\circ\) to \(42^\circ\)). This smaller angle provides a higher horizontal velocity component to offset the cumulative drag deceleration, while spending slightly less time in the air where drag acts.
Curriculum Framework: AP Physics 1 Subject: Experimental Design Questions Key
