Projectile motion
Problem 9.759 · medium
A projectile is fired from \( \displaystyle 2 \) 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) = 2 + v₀sin θ · t − ½gt².Reviewed
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 2\right) \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} }} = 0 \]The vertical velocity is 0 at t = 50*sqrt(2)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 2 \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} }} = \frac{598}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{2} t + 2 \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} + \frac{2 \sqrt{1495}}{49} }} = 0 \]It lands (y = 0) at t = 50*sqrt(2)/49 + 2*sqrt(1495)/49.✓ Proved
- \[ \left. 10 \sqrt{2} t \right|_{\substack{ t=\frac{50 \sqrt{2}}{49} + \frac{2 \sqrt{1495}}{49} }} = \frac{1000}{49} + \frac{20 \sqrt{2990}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{598}{49} \approx 12.20,\ T = \frac{50 \sqrt{2}}{49} + \frac{2 \sqrt{1495}}{49} \approx 3.021,\ \text{range} = \frac{1000}{49} + \frac{20 \sqrt{2990}}{49} \approx 42.73 \)
✓ 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
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-18gpt-oss:20b: pass 2026-10-18qwen3.6:27b-mlx: pass 2026-10-16gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15gpt-oss:20b: pass 2026-10-15
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 2026-10-18 with SymPy 1.14.0.