Projectile motion
Problem 9.896 · medium
A projectile is fired from \( \displaystyle 1 \) ft above the ground at \( \displaystyle 64 \) ft/s, \( \displaystyle 60^\circ \) 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.
- \[ \left. \frac{d}{d t} \left(- 16 t^{2} + 32 \sqrt{3} t + 1\right) \right|_{\substack{ t=\sqrt{3} }} = 0 \]The vertical velocity is 0 at t = sqrt(3).✓ Proved
- \[ \left. - 16 t^{2} + 32 \sqrt{3} t + 1 \right|_{\substack{ t=\sqrt{3} }} = 49 \]The maximum height.✓ Proved
- \[ \left. - 16 t^{2} + 32 \sqrt{3} t + 1 \right|_{\substack{ t=\sqrt{3} + \frac{7}{4} }} = 0 \]It lands (y = 0) at t = sqrt(3) + 7/4.✓ Proved
- \[ \left. 32 t \right|_{\substack{ t=\sqrt{3} + \frac{7}{4} }} = 32 \sqrt{3} + 56 \]The horizontal distance then.✓ Proved
Answer \( H = 49 \approx 49.00,\ T = \sqrt{3} + \frac{7}{4} \approx 3.482,\ \text{range} = 32 \sqrt{3} + 56 \approx 111.43 \)
Lines: 4 proved, 1 not checked. 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 | Not checked | — | 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: fail (style) — [domain objection, downgraded to style] The horizontal position function x(t) is incorrectly defined as 32t in step 5, omitting the cos(60°) factor. The initial horizontal velocity is 32 ft/s, not 64 ft/s, so the range calculation is double the correct value.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-11-06 — [domain objection, downgraded to style] The horizontal position function x(t) is incorrectly defined as 32t in step 5, omitting the cos(60°) factor. The initial horizontal velocity is 32 ft/s, not 64 ft/s, so the range calculation is double the correct value.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: fail (error) 2026-11-05 — The horizontal position function in line 1 is incorrect; it uses 32t instead of the correct 32\sqrt{3}t (since v_x = 64 \cos(60^\circ) = 32). Consequently, the range calculation in line 5 is wrong.gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: fail (error) 2026-11-05 — The horizontal position function x(t) in line 1 is incorrect; it uses v₀ instead of v₀cos(θ). Consequently, the horizontal range calculation in line 5 is wrong (it calculates distance as if the horizontal velocity were 64 ft/s instead of 32 ft/s).gpt-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.