Projectile motion
Problem 9.777 · medium
A projectile is fired from \( \displaystyle 2 \) m above the ground at \( \displaystyle 40 \) 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) = 2 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 20 \sqrt{3} t + 2\right) \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 100*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 20 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} }} = \frac{3098}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 20 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} + \frac{2 \sqrt{7745}}{49} }} = 0 \]It lands (y = 0) at t = 100*sqrt(3)/49 + 2*sqrt(7745)/49.✓ Proved
- \[ \left. 20 t \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} + \frac{2 \sqrt{7745}}{49} }} = \frac{2000 \sqrt{3}}{49} + \frac{40 \sqrt{7745}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{3098}{49} \approx 63.22,\ T = \frac{100 \sqrt{3}}{49} + \frac{2 \sqrt{7745}}{49} \approx 7.127,\ \text{range} = \frac{2000 \sqrt{3}}{49} + \frac{40 \sqrt{7745}}{49} \approx 142.54 \)
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 in line 1 is defined as x(t) = v₀cos θ · t, which evaluates to 20√3 t. However, line 5 substitutes 20*t into the position function, omitting the √3 factor from the horizontal velocity component. This results in an incorrect range calculation.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (style) 2026-10-19 — [domain objection, downgraded to style] The horizontal position function in line 1 is defined as x(t) = v₀cos θ · t, which evaluates to 20√3 t. However, line 5 substitutes 20*t into the position function, omitting the √3 factor from the horizontal velocity component. This results in an incorrect range calculation.gpt-oss:20b: pass 2026-10-19qwen3.6:27b-mlx: fail (error) 2026-10-16 — The horizontal velocity component is v₀cos(60°) = 20 m/s, but the solution incorrectly uses 20t for the horizontal position in step 5, implying a horizontal velocity of 20 m/s. Wait, v₀=40, cos(60)=0.5, so v_x = 20. The equation x(t) = 20t is correct. Let me re-read carefully. Step 1 says x(t) = v₀cos θ · t. Step 5 substitutes into 20*t. This is correct. Why did I think it was wrong? Ah, I need to check the vertical equation. y(t) = 2 + 40sin(60)t - 4.9t^2 = 2 + 20√3 t - 4.9t^2. Step 2 uses -49t^2/10 + 20√3t + 2. This is correct. Step 3 calculates max height. Step 4 calculates time of flight. Step 5 calculates range. The solution seems algebraically correct. Let me check the numerical values. H = 3098/49 ≈ 63.22. T ≈ 7.127. Range ≈ 142.54. The logic holds. Is there a subtle error? The problem asks for max height, time of flight, and horizontal range. The solution provides them. The setup is correct. The equations are correct. The substitutions are correct. I see no error.gpt-oss:20b: pass 2026-10-16qwen3.6:27b-mlx: inconclusive 2026-10-15 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal velocity component is incorrect. With v₀=40 and θ=60°, v_x should be 20, but the solution uses 20*sqrt(3) (which is v_y). Consequentlygpt-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.