Projectile motion
Problem 9.826 · medium
A projectile is fired from \( \displaystyle 2 \) m above the ground at \( \displaystyle 80 \) m/s, \( \displaystyle 30^\circ \) 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
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 40 t + 2\right) \right|_{\substack{ t=\frac{200}{49} }} = 0 \]The vertical velocity is 0 at t = 200/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 t + 2 \right|_{\substack{ t=\frac{200}{49} }} = \frac{4098}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 t + 2 \right|_{\substack{ t=\frac{200}{49} + \frac{2 \sqrt{10245}}{49} }} = 0 \]It lands (y = 0) at t = 200/49 + 2*sqrt(10245)/49.✓ Proved
- \[ \left. 40 \sqrt{3} t \right|_{\substack{ t=\frac{200}{49} + \frac{2 \sqrt{10245}}{49} }} = \frac{80 \sqrt{3} \left(100 + \sqrt{10245}\right)}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{4098}{49} \approx 83.63,\ T = \frac{200}{49} + \frac{2 \sqrt{10245}}{49} \approx 8.213,\ \text{range} = \frac{80 \sqrt{3} \left(100 + \sqrt{10245}\right)}{49} \approx 569.01 \)
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
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-10-20gpt-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: fail (error) 2026-10-19 — The time of flight is incorrectly computed. The quadratic y(t)=0 has roots t≈4.08 s (ascending) and t≈8.21 s (descending). The solution uses an incorrect root (≈6.15 s), leading to an incorrect horizontal range. The setup and earlier steps are fine, but the application of the quadratic formula in step 4 is wrong.qwen3.6:27b-mlx: pass 2026-10-19 — The solution correctly models the projectile motion with initial height, identifies the time of maximum height via zero vertical velocity, solves for the landing time, and computes the range. The algebraic steps are consistent with the stated answer.gpt-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.