Projectile motion
Problem 9.684 · medium
A projectile is fired from \( \displaystyle 1 \) m above the ground at \( \displaystyle 50 \) 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.
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 25 \sqrt{3} t + 1\right) \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 125*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 25 \sqrt{3} t + 1 \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} }} = \frac{9473}{98} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 25 \sqrt{3} t + 1 \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} + \frac{\sqrt{47365}}{49} }} = 0 \]It lands (y = 0) at t = 125*sqrt(3)/49 + sqrt(47365)/49.✓ Proved
- \[ \left. 25 t \right|_{\substack{ t=\frac{125 \sqrt{3}}{49} + \frac{\sqrt{47365}}{49} }} = \frac{3125 \sqrt{3}}{49} + \frac{25 \sqrt{47365}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{9473}{98} \approx 96.66,\ T = \frac{125 \sqrt{3}}{49} + \frac{\sqrt{47365}}{49} \approx 8.860,\ \text{range} = \frac{3125 \sqrt{3}}{49} + \frac{25 \sqrt{47365}}{49} \approx 221.50 \)
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 (error) — The horizontal position function in line 1 is incorrect; it uses 25t instead of 25\sqrt{3}t, leading to an incorrect range calculation in line 5.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The horizontal position function in line 1 is incorrect; it uses 25t instead of 25\sqrt{3}t, leading to an incorrect range calculation in line 5.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The horizontal position function in line 1 is incorrect; it uses 25t instead of 25\sqrt{3}t (since v_0 \cos(60^\circ) = 25). Consequently, the calculated range in line 5 is half the correct value.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: fail (error) 2026-10-14 — The horizontal velocity component is v₀cos(60°) = 25 m/s, but the solution incorrectly uses x(t) = 25t in step 5 while the vertical equation in step 2 implies a different scaling or is inconsistent with standard projectile motion formulas (specifically, the coefficient of t² in y(t) should be -4.9, not -4.9/10 as implied by the CAS check if g=9.8, wait: -1/2 * 9.8 = -4.9. The CAS check shows -49/10 = -4.9, so that is correct. However, step 1 states x(t) = v₀cos θ · t. v₀=50, cos(60)=0.5, so v_x = 25. Step 5 uses 25*t. This is consistent. Let's re-read carefully. Step 2 checks derivative of y(t). y(t) = 1 + 25√3 t - 4.9 t². Derivative is 25√3 - 9.8t. Setting to 0 gives t = 25√3 / 9.8 = 250√3 / 98 = 125√3 / 49. This is correct. Step 3 calculates max height. Step 4 finds time of flight. Step 5 calculates range. The algebra seems correct. Why did I think it was wrong? Let me check the 'unchecked' equations. The prompt says 'equation, unchecked' but then provides a CAS result. The prompt says 'do not re-check the algebra of a line marked proved'. These are marked 'unchecked' but have CAS results. I should trust the CAS results for algebra. Is there a conceptual error? The problem asks for max height, time of flight, horizontal range. The solution provides them. The setup in step 1 is correct. The calculations in 2-5 are verified by CAS. Wait, look at step 5. x(t) = 25t. The range is x(T). The CAS check for step 5 substitutes T into 25t. This is correct. Is there any subtle error? The problem states 'find the maximum height, the time of flight and the horizontal range'. The solution provides exact and approximate values. The approximations match the exact values. 9473/98 ≈ 96.66. 125√3/49 + √47365/49 ≈ 4.33 + 4.53 = 8.86. Range ≈ 25 * 8.86 = 221.5. Everything seems correct. Why would it fail? Let me look closer at the prompt's instruction: 'Each line... is either an EQUATION... or a SENTENCE'. Step 1 is a sentence. Steps 2-5 are equations. I only review sentences? 'The sentences are what you are here for.' Step 1 is the only sentence. 'x(t) = v₀cos θ · t and y(t) = 1 + v₀sin θ · t − ½gt².' This is a correct model. Is there a missing sentence? The solution jumps from the model to the CAS checks. It doesn't explicitly state 'The maximum height occurs when vertical velocity is zero' or 'The time of flight is when y(t)=0'. However, the comments in the equation lines explain this. The prompt says 'Check that... each conclusion follows from the lines before it'. The conclusions (the values) are in the 'Stated answer', not in the solution lines themselves? No, the solution lines *are* the derivation. The comments '-- The maximum height' etc are part of the equation lines? Or are they separate? The format is `2. [equation, unchecked] ... -- comment`. The comment is likely part of the line. If I only review sentences, and step 1 is the only sentence, and it is correct, then it passes? But usually, these adversarial reviews catch missing logical steps or incorrect setups. Step 1 is correct. Is there a hidden error? g=9.8. 1/2 g = 4.9. 4.9 = 49/10. Correct. v0=50. sin(60)=√3/2. v0y = 25√3. Correct. cos(60)=1/2. v0x = 25. Correct. Initial height y0=1. Correct. The solution is sound. I will mark it pass.gpt-oss:20b: pass 2026-10-14
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 2026-10-17 with SymPy 1.14.0.