Projectile motion
Problem 9.672 · medium
A projectile is fired from \( \displaystyle 10 \) m above the ground at \( \displaystyle 20 \) 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) = 10 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 10 \sqrt{3} t + 10\right) \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 50*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{3} t + 10 \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} }} = \frac{1240}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{3} t + 10 \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} + \frac{20 \sqrt{31}}{49} }} = 0 \]It lands (y = 0) at t = 50*sqrt(3)/49 + 20*sqrt(31)/49.✓ Proved
- \[ \left. 10 t \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} + \frac{20 \sqrt{31}}{49} }} = \frac{500 \sqrt{3}}{49} + \frac{200 \sqrt{31}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{1240}{49} \approx 25.31,\ T = \frac{50 \sqrt{3}}{49} + \frac{20 \sqrt{31}}{49} \approx 4.040,\ \text{range} = \frac{500 \sqrt{3}}{49} + \frac{200 \sqrt{31}}{49} \approx 40.40 \)
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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function in line 1 is defined as x(t) = v₀cos θ · t, but line 5 incorrectly substitutes 10*t into the equation to calculate t
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function in line 1 is defined as x(t) = v₀cos θ · t, but line 5 incorrectly substitutes 10*t into the equation to calculate tgpt-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 line 1 is defined as x(t) = v₀cos θ · t, but line 5 incorrectly substitutes 10*t into the equation for horizontagpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The setup in line 1 is correct, but the subsequent equations use an incorrect y(t) formula (-49/10 t^2 instead of -4.9 t^2) and an incorrect x(t) formula (10t instead of 10*sqrt(3)t). This leads to wrong numerical results for time of flight and range.gpt-oss:20b: fail (error) 2026-10-14 — The time of flight is incorrectly computed; the quadratic solution for y=0 gives t=50√3/49 – 20√31/49, not plus. The subsequent range calculation is therefore also incorrect.
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.