Projectile motion
Problem 9.895 · medium
A projectile is fired from \( \displaystyle 0 \) m above the ground at \( \displaystyle 50 \) 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.
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 25 \sqrt{3} t\right) \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 125*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 25 \sqrt{3} t \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} }} = \frac{9375}{98} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 25 \sqrt{3} t \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = 0 \]It lands (y = 0) at t = 250*sqrt(3)/49.✓ Proved
- \[ \left. 25 t \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = \frac{6250 \sqrt{3}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{9375}{98} \approx 95.66,\ T = \frac{250 \sqrt{3}}{49} \approx 8.837,\ \text{range} = \frac{6250 \sqrt{3}}{49} \approx 220.92 \)
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 step 1 is incorrect; it uses x(t) = v₀t instead of x(t) = v₀cos(θ)t. This error is masked in step 5 because the specific value of v₀cos(60°) happens to be 25, but the stated formula is mathematically false and would teach a student incorrect physics.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The horizontal position function in step 1 is incorrect; it uses x(t) = v₀t instead of x(t) = v₀cos(θ)t. This error is masked in step 5 because the specific value of v₀cos(60°) happens to be 25, but the stated formula is mathematically false and would teach a student incorrect physics.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: fail (error) 2026-11-05 — The horizontal position function in step 1 is incorrect; it uses x(t) = 25t instead of x(t) = 25\sqrt{3}t, omitting the \cos(60^\circ) factor. Consequently, the range calculation in step 5 is wrong.gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: fail (error) 2026-11-05 — The solution fails to substitute the given values (v₀=50, θ=60°) into the general formulas in step 1. Consequently, the subsequent calculations use incorrect coefficients (e.g., 25 instead of 25√3 for horizontal velocity), leading to wrong numerical answers.gpt-oss:20b: pass 2026-11-05
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 2026-11-06 with SymPy 1.14.0.