Projectile motion
Problem 9.674 · medium
A projectile is fired from \( \displaystyle 0 \) m above the ground at \( \displaystyle 80 \) m/s, \( \displaystyle 60^\circ \) 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) = 0 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 40 \sqrt{3} t\right) \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 200*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{3} t \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} }} = \frac{12000}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{3} t \right|_{\substack{ t=\frac{400 \sqrt{3}}{49} }} = 0 \]It lands (y = 0) at t = 400*sqrt(3)/49.✓ Proved
- \[ \left. 40 t \right|_{\substack{ t=\frac{400 \sqrt{3}}{49} }} = \frac{16000 \sqrt{3}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{12000}{49} \approx 244.90,\ T = \frac{400 \sqrt{3}}{49} \approx 14.139,\ \text{range} = \frac{16000 \sqrt{3}}{49} \approx 565.57 \)
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 horizontal position function in line 1 is missing the cosine component (v₀ cos θ), leading to an incorrect range calculation in line 5. The solution uses x(t) = 40t instead of x(t) = 40√3 t.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The horizontal position function in line 1 is missing the cosine component (v₀ cos θ), leading to an incorrect range calculation in line 5. The solution uses x(t) = 40t instead of x(t) = 40√3 t.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function in step 1 is defined as x(t) = v₀cos θ · t, but step 5 substitutes 40*t into the equation. Since v₀cos(60°) = 40, thgpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The solution uses an incorrect horizontal velocity component (40 m/s instead of 40\sqrt{3} m/s) in the final calculation for the range, leading to an incorrect result.gpt-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.