Projectile motion
Problem 9.785 · medium
A projectile is fired from \( \displaystyle 1 \) m above the ground at \( \displaystyle 100 \) m/s, \( \displaystyle 60^\circ \) 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) = 1 + v₀sin θ · t − ½gt².Reviewed
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 1\right) \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 250*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 1 \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = \frac{18799}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 1 \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} + \frac{\sqrt{187990}}{49} }} = 0 \]It lands (y = 0) at t = 250*sqrt(3)/49 + sqrt(187990)/49.✓ Proved
- \[ \left. 50 t \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} + \frac{\sqrt{187990}}{49} }} = \frac{12500 \sqrt{3}}{49} + \frac{50 \sqrt{187990}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{18799}{49} \approx 383.65,\ T = \frac{250 \sqrt{3}}{49} + \frac{\sqrt{187990}}{49} \approx 17.686,\ \text{range} = \frac{12500 \sqrt{3}}{49} + \frac{50 \sqrt{187990}}{49} \approx 884.28 \)
✓ 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 initial height, accurately identifies the time of maximum height and landing time, and computes the range. The algebraic steps are verified and consistent with the stated answer.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly models the projectile motion with initial height, accurately identifies the time of maximum height and landing time, and computes the range. The algebraic steps are verified and consistent with the stated answer.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: pass 2026-10-16 — The solution correctly models the projectile motion with initial height, accurately identifies the time of maximum height and landing time, and computes the range correctly.gpt-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 2026-10-19 with SymPy 1.14.0.