Projectile motion
Problem 9.828 · medium
A projectile is fired from \( \displaystyle 2 \) m above the ground at \( \displaystyle 100 \) 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
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 50 \sqrt{2} t + 2\right) \right|_{\substack{ t=\frac{250 \sqrt{2}}{49} }} = 0 \]The vertical velocity is 0 at t = 250*sqrt(2)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{2} t + 2 \right|_{\substack{ t=\frac{250 \sqrt{2}}{49} }} = \frac{12598}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{2} t + 2 \right|_{\substack{ t=\frac{250 \sqrt{2}}{49} + \frac{2 \sqrt{31495}}{49} }} = 0 \]It lands (y = 0) at t = 250*sqrt(2)/49 + 2*sqrt(31495)/49.✓ Proved
- \[ \left. 50 \sqrt{2} t \right|_{\substack{ t=\frac{250 \sqrt{2}}{49} + \frac{2 \sqrt{31495}}{49} }} = \frac{25000}{49} + \frac{100 \sqrt{62990}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{12598}{49} \approx 257.10,\ T = \frac{250 \sqrt{2}}{49} + \frac{2 \sqrt{31495}}{49} \approx 14.459,\ \text{range} = \frac{25000}{49} + \frac{100 \sqrt{62990}}{49} \approx 1022.40 \)
✓ 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 models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range using standard kinematic equations.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-20 — The solution correctly models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range using standard kinematic equations.gpt-oss:20b: pass 2026-10-20qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-19gpt-oss:20b: pass 2026-10-19
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 2026-10-20 with SymPy 1.14.0.