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Horizontal Projectiles

A comprehensive physics lesson exploring two-dimensional kinematics, horizontal projectile motion, and the determination of gravitational acceleration via PhET simulation experiments and mathematical modeling.

CMChuguna Murugesan

Horizontal Projectiles

A comprehensive physics lesson exploring two-dimensional kinematics, horizontal projectile motion, and the determination of gravitational acceleration via PhET simulation experiments and mathematical modeling.

CMChuguna Murugesan
Controlled Variable 2Launch Angle (\(\theta = 0^\circ\)), Air Drag (OFF)Launch Angle (\(\theta = 0^\circ\)), Air Drag (OFF)
Unmeasured VariableTime of Flight (\(t\))Time of Flight (\(t\))

Physics Instructor Guide: Horizontal Projectile Kinematics Answer Key — Page 1 of 2

Teacher Answer Key | Post-Lab Synthesis: Solutions & Grading Criteria

Grading Guide

Part A: Launch Velocity (\(v_0\)) vs. Range (\(x\)) Solutions

1. Proportionality (\(x \propto v_0\)):

Answer: Yes. The data exhibits direct linear proportionality. Doubling launch speed from \(10\text{ m/s}\) to \(20\text{ m/s}\) exactly doubles the horizontal range (e.g., from \(\sim 14.3\text{ m}\) to \(\sim 28.6\text{ m}\) at \(h = 10\text{ m}\)). Factor increase = \(2\times\).

2. Fall Time Invariance:

Answer: Flight time remained completely unchanged across all velocity trials. Since initial vertical velocity is zero and vertical acceleration is \(g\), flight time is strictly \(t = \sqrt{2h/g}\). This directly validates Galileo's assertion that orthogonal kinematic motions operate independently.

Part B: Elevation Height (\(h\)) vs. Range (\(x\)) Solutions

3. Non-Linear Scaling (\(x \propto \sqrt{h}\)):

Answer: The range did not quadruple; it only doubled (\(2\times\)). Because range depends on flight time and \(t \propto \sqrt{h}\), increasing height by a factor of 4 increases range by a factor of \(\sqrt{4} = 2\).

4. Vertical Drop & Flight Time:

Answer: Yes, flight time changed significantly. Unlike horizontal speed, elevating the launch position increases the vertical distance the object must free-fall to hit the ground.

Part C: Calculating Gravity from Part 2 Data Only Solution

5. Solution for Calculating \(g\) from Part 2 Measurements: \(g = \frac{2h}{t^2} = \frac{2h v_0^2}{x^2}\)

Step 1 (Identify knowns from Part 2): Cannon elevation height \(h\) (known & varied), launch speed \(v_0\) (known constant), and ground impact range \(x\) (measured with tape).

Step 2 (Find flight time \(t\)): Because horizontal speed is constant (\(a_x = 0\)), \(x = v_0 t \implies t = \frac{x}{v_0}\).

Step 3 (Apply vertical kinematic equation): Free fall from rest gives \(h = \frac{1}{2}gt^2 \implies g = \frac{2h}{t^2}\).

Step 4 (Substitute \(t\)): Substituting \(t = \frac{x}{v_0}\) into \(g = \frac{2h}{t^2}\) yields \(g = \frac{2h v_0^2}{x^2}\). Students receive full credit for stating \(g = \frac{2h}{t^2}\) (explaining how \(t = x/v_0\) is found) or writing the full substituted form.

Part D: Real-World Air Resistance Solutions

Horizontal Velocity Decay:

Horizontal velocity does not stay constant. Aerodynamic drag exerts an opposing force (\(F_{dx} \propto v_x^2\)), producing a negative horizontal acceleration (\(a_x < 0\)). Consequently, horizontal range is noticeably reduced.

Trajectory Asymmetry:

The trajectory ceases to be a true parabola. As horizontal velocity decays while downward speed continues to accelerate toward terminal velocity, the path steepens dramatically into an asymmetric dive near the ground.

Holistic Grading Rubric (Total: 25 Points) 100% Scale

Pre-Lab Q1–Q3 6 pts (2 pts each)

Formulas & Table 7 pts (4 pts + 3 pts)

Post-Lab Part A & B 6 pts (1.5 pts each)

Part C & D (g & Drag) 6 pts (3 pts each)

Physics Mechanics: Post-Laboratory Synthesis & Evaluation Answer Key — Page 2 of 2

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