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
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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
