Projectile motion
Problem 9.894 · medium
A projectile is fired from \( \displaystyle 0 \) m above the ground at \( \displaystyle 100 \) m/s, \( \displaystyle 30^\circ \) 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.Reviewed
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 50 t\right) \right|_{\substack{ t=\frac{250}{49} }} = 0 \]The vertical velocity is 0 at t = 250/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 t \right|_{\substack{ t=\frac{250}{49} }} = \frac{6250}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 t \right|_{\substack{ t=\frac{500}{49} }} = 0 \]It lands (y = 0) at t = 500/49.✓ Proved
- \[ \left. 50 \sqrt{3} t \right|_{\substack{ t=\frac{500}{49} }} = \frac{25000 \sqrt{3}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{6250}{49} \approx 127.55,\ T = \frac{500}{49} \approx 10.204,\ \text{range} = \frac{25000 \sqrt{3}}{49} \approx 883.70 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. 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 flight and maximum height via standard calculus/algebraic methods, and computes the range accurately. The intermediate coefficients (e.g., 49/10 for 1/2 g) are correct.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-06 — The solution correctly sets up the kinematic equations, identifies the time of flight and maximum height via standard calculus/algebraic methods, and computes the range accurately. The intermediate coefficients (e.g., 49/10 for 1/2 g) are correct.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: pass 2026-11-05gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: pass 2026-11-05gpt-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.