Projectile motion
Problem 9.898 · medium
A projectile is fired from \( \displaystyle 10 \) ft above the ground at \( \displaystyle 32 \) 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) = 10 + v₀sin θ · t − ½gt².Reviewed
- \[ \left. \frac{d}{d t} \left(- 16 t^{2} + 16 \sqrt{2} t + 10\right) \right|_{\substack{ t=\frac{\sqrt{2}}{2} }} = 0 \]The vertical velocity is 0 at t = sqrt(2)/2.✓ Proved
- \[ \left. - 16 t^{2} + 16 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{\sqrt{2}}{2} }} = 18 \]The maximum height.✓ Proved
- \[ \left. - 16 t^{2} + 16 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{5 \sqrt{2}}{4} }} = 0 \]It lands (y = 0) at t = 5*sqrt(2)/4.✓ Proved
- \[ \left. 16 \sqrt{2} t \right|_{\substack{ t=\frac{5 \sqrt{2}}{4} }} = 40 \]The horizontal distance then.✓ Proved
Answer \( H = 18 \approx 18.00,\ T = \frac{5 \sqrt{2}}{4} \approx 1.768,\ \text{range} = 40 \approx 40.00 \)
Lines: 4 proved, 1 reviewed. 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 | 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 initial height, derives the correct time of flight and maximum height, and calculates the range accurately.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-06 — The solution correctly models the projectile motion with initial height, derives the correct time of flight and maximum height, and calculates the range accurately.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: pass 2026-11-05 — 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 2026-11-05qwen3.6:27b-mlx: pass 2026-11-05gpt-oss:20b: fail (error) 2026-11-05 — The time of flight is incorrectly computed; solving y(t)=0 gives t=1.25 s, not 5√2/4≈1.768 s. The horizontal range should be 16√2·1.25≈25 ft, not 40 ft.
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.