Projectile motion
Problem 9.685 · medium
A projectile is fired from \( \displaystyle 10 \) m above the ground at \( \displaystyle 20 \) 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.Reviewed
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 10\right) \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} }} = 0 \]The vertical velocity is 0 at t = 50*sqrt(2)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} }} = \frac{990}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} + \frac{30 \sqrt{11}}{49} }} = 0 \]It lands (y = 0) at t = 50*sqrt(2)/49 + 30*sqrt(11)/49.✓ Proved
- \[ \left. 10 \sqrt{2} t \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} + \frac{30 \sqrt{11}}{49} }} = \frac{1000}{49} + \frac{300 \sqrt{22}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{990}{49} \approx 20.20,\ T = \frac{50 \sqrt{2}}{49} + \frac{30 \sqrt{11}}{49} \approx 3.474,\ \text{range} = \frac{1000}{49} + \frac{300 \sqrt{22}}{49} \approx 49.12 \)
Lines: 4 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly sets up the kinematic equations, identifies the time of maximum height via zero vertical velocity, solves for the landing time, and computes the range. All steps are logically sound and algebraically verified.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-17 — The solution correctly sets up the kinematic equations, identifies the time of maximum height via zero vertical velocity, solves for the landing time, and computes the range. All steps are logically sound and algebraically verified.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: pass 2026-10-15 — The solution correctly sets up the kinematic equations, identifies the time of maximum height via zero vertical velocity, solves for the landing time, and computes the range. All steps are logically sound and algebraically verified.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The solution uses an incorrect equation for vertical position y(t) in steps 2-4, using -49/10 instead of -49/2 (or -g/2). This leads to incorrect vagpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.