Projectile motion
Problem 9.825 · medium
A projectile is fired from \( \displaystyle 10 \) m above the ground at \( \displaystyle 80 \) m/s, \( \displaystyle 45^\circ \) above the horizontal. Taking \( \displaystyle g = 9.8 \) m/s², find the maximum height, the time of flight and the horizontal range.
- x(t) = v₀cos θ · t and y(t) = 10 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 40 \sqrt{2} t + 10\right) \right|_{\substack{ t=\frac{200 \sqrt{2}}{49} }} = 0 \]The vertical velocity is 0 at t = 200*sqrt(2)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{200 \sqrt{2}}{49} }} = \frac{8490}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{200 \sqrt{2}}{49} + \frac{10 \sqrt{849}}{49} }} = 0 \]It lands (y = 0) at t = 200*sqrt(2)/49 + 10*sqrt(849)/49.✓ Proved
- \[ \left. 40 \sqrt{2} t \right|_{\substack{ t=\frac{200 \sqrt{2}}{49} + \frac{10 \sqrt{849}}{49} }} = \frac{16000}{49} + \frac{400 \sqrt{1698}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{8490}{49} \approx 173.27,\ T = \frac{200 \sqrt{2}}{49} + \frac{10 \sqrt{849}}{49} \approx 11.719,\ \text{range} = \frac{16000}{49} + \frac{400 \sqrt{1698}}{49} \approx 662.91 \)
Lines: 4 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 4 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 5 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | the flight stepped forward in time steps of 10⁻⁴ s |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly uses g = 4.9 m/s² (evident from the coefficient -49/10 = -4.9 in the equations) instead of the stated g = 9.8 m/s². This results in incorrect values for maximum height, time of flight, and range.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-20 — The solution incorrectly uses g = 4.9 m/s² (evident from the coefficient -49/10 = -4.9 in the equations) instead of the stated g = 9.8 m/s². This results in incorrect values for maximum height, time of flight, and range.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range using standard kinematic equations.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-19 — The solution uses an incorrect coefficient for the quadratic term in the y(t) equation (-49/10 instead of -4.9), which contradicts the stated g=9.8 m/s² and leads to incorrect numerical results for time and range.gpt-oss:20b: pass 2026-10-19
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/projectile_motion, checked 2026-10-20 with SymPy 1.14.0.