Projectile motion
Problem 9.782 · medium
A projectile is fired from \( \displaystyle 1 \) ft above the ground at \( \displaystyle 96 \) ft/s, \( \displaystyle 45^\circ \) above the horizontal. Taking \( \displaystyle g = 32 \) ft/s², find the maximum height, the time of flight and the horizontal range.
- x(t) = v₀cos θ · t and y(t) = 1 + v₀sin θ · t − ½gt².Reviewed
- \[ \left. \frac{d}{d t} \left(- 16 t^{2} + 48 \sqrt{2} t + 1\right) \right|_{\substack{ t=\frac{3 \sqrt{2}}{2} }} = 0 \]The vertical velocity is 0 at t = 3*sqrt(2)/2.✓ Proved
- \[ \left. - 16 t^{2} + 48 \sqrt{2} t + 1 \right|_{\substack{ t=\frac{3 \sqrt{2}}{2} }} = 73 \]The maximum height.✓ Proved
- \[ \left. - 16 t^{2} + 48 \sqrt{2} t + 1 \right|_{\substack{ t=\frac{3 \sqrt{2}}{2} + \frac{\sqrt{73}}{4} }} = 0 \]It lands (y = 0) at t = 3*sqrt(2)/2 + sqrt(73)/4.✓ Proved
- \[ \left. 48 \sqrt{2} t \right|_{\substack{ t=\frac{3 \sqrt{2}}{2} + \frac{\sqrt{73}}{4} }} = 144 + 12 \sqrt{146} \]The horizontal distance then.✓ Proved
Answer \( H = 73 \approx 73.00,\ T = \frac{3 \sqrt{2}}{2} + \frac{\sqrt{73}}{4} \approx 4.257,\ \text{range} = 144 + 12 \sqrt{146} \approx 289.00 \)
✓ 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.
Every verdict on record (6)
qwen3.6:27b-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 2026-10-16 — 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 2026-10-16qwen3.6:27b-mlx: pass 2026-10-15 — 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 2026-10-15
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 2026-10-19 with SymPy 1.14.0.