Projectile motion
Problem 9.892 · medium
A projectile is fired from \( \displaystyle 2 \) 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) = 2 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 2\right) \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 250*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} }} = \frac{18848}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 50 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} + \frac{8 \sqrt{2945}}{49} }} = 0 \]It lands (y = 0) at t = 250*sqrt(3)/49 + 8*sqrt(2945)/49.✓ Proved
- \[ \left. 50 t \right|_{\substack{ t=\frac{250 \sqrt{3}}{49} + \frac{8 \sqrt{2945}}{49} }} = \frac{12500 \sqrt{3}}{49} + \frac{400 \sqrt{2945}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{18848}{49} \approx 384.65,\ T = \frac{250 \sqrt{3}}{49} + \frac{8 \sqrt{2945}}{49} \approx 17.697,\ \text{range} = \frac{12500 \sqrt{3}}{49} + \frac{400 \sqrt{2945}}{49} \approx 884.85 \)
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: fail (error) — The time of impact is computed incorrectly: the quadratic solution should give t = (500√3 – 16√2945)/98, not (500√3 + 16√2945)/98. This propagates to an incorrect range.qwen3.6:27b-mlx: fail (error) — The horizontal position function in line 5 is incorrect; it uses 50t instead of 50\sqrt{3}t, omitting the \cos(60^\circ) factor. This leads to an incorrect range calculation.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The horizontal position function in line 5 is incorrect; it uses 50t instead of 50\sqrt{3}t, omitting the \cos(60^\circ) factor. This leads to an incorrect range calculation.gpt-oss:20b: fail (error) 2026-11-06 — The time of impact is computed incorrectly: the quadratic solution should give t = (500√3 – 16√2945)/98, not (500√3 + 16√2945)/98. This propagates to an incorrect range.qwen3.6:27b-mlx: fail (error) 2026-11-05 — The horizontal position function in line 5 is incorrect; it uses 50t instead of 50\sqrt{3}t, omitting the \cos(60^\circ) factor. This leads to an incorrect range calculation.gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: fail (error) 2026-11-05 — The horizontal velocity component is calculated as 50 m/s (v₀ cos 60°), but the range calculation in step 5 uses this value directly. However, the time of flight T is derived from the quadratic formula for y(t)=0. The expression for T in step 4 is correct for the roots of the quadratic, but the range calculation in step 5 substitutes this T into x(t) = 50t. The algebraic simplification of the range is correct based on the T found. Wait, let's re-read carefully. Step 1 defines x(t) and y(t). Step 2 finds t_max. Step 3 finds H. Step 4 finds T (landing time). Step 5 finds Range. The issue is that the problem asks for 'horizontal range'. The solution provides the correct numerical values. Is there a subtle error? Let's check the quadratic solution. y(t) = -4.9t^2 + 50sqrt(3)t + 2 = 0. Roots are [-50sqrt(3) +/- sqrt(75000 + 39.2)] / -9.8. sqrt(75392) = sqrt(16 * 4712) = 4 sqrt(4712) = 4 sqrt(4 * 1178) = 8 sqrt(1178). 1178 = 2 * 589. 589 is 19*31. So sqrt(75392) = 8 sqrt(1178). The solution uses sqrt(2945). 2945 * 4 = 11780. No. 2945 * 25.6? Let's check 49^2 * 2945 / 100? No. Let's check the discriminant again. b^2 - 4ac = (50sqrt(3))^2 - 4(-4.9)(2) = 7500 + 39.2 = 7539.2. sqrt(7539.2) = sqrt(75392/10) = sqrt(37696/5). The solution has 8*sqrt(2945)/49. (8*sqrt(2945)/49)^2 = 64*2945/2401 = 188480/2401 ≈ 78.5. This does not match the discriminant part of the quadratic formula directly. The time T is given as 250sqrt(3)/49 + 8sqrt(2945)/49. Let's check if this is a root. t = [ -b +/- sqrt(D) ] / 2a. Here a = -4.9 = -49/10. b = 50sqrt(3). c = 2. t = [ -50sqrt(3) +/- sqrt(7539.2) ] / -9.8. t = [ 50sqrt(3) -/ sqrt(7539.2) ] / 9.8. 50sqrt(3)/9.8 = 500sqrt(3)/98 = 250sqrt(3)/49. This matches the first term. The second term is sqrt(7539.2)/9.8. sqrt(7539.2) = sqrt(75392/10). 75392 = 16 * 4712 = 64 * 1178. So sqrt(7539.2) = 8 sqrt(117.8). 117.8 = 1178/10. So 8 sqrt(1178/10) / 9.8. The solution has 8 sqrt(2945) / 49. Let's square 8 sqrt(2945) / 49: 64 * 2945 / 2401 = 188480 / 2401 ≈ 78.5. Let's square sqrt(7539.2)/9.8: 7539.2 / 96.04 ≈ 78.5. They are numerically equal. 2945 * 64 = 188480. 7539.2 * 25 = 188480. So 2945 * 64 = 7539.2 * 25? 188480 = 188480. Yes. So the algebra is correct. Why would it fail? The prompt says 'find the maximum height, the time of flight and the horizontal range'. The solution provides these. Is there a sign error? No. Is the setup wrong? No. Wait, look at step 1. x(t) = v0 cos theta t. y(t) = 2 + v0 sin theta t - 0.5 g t^2. This is correct. Step 2: Derivative of y(t) is -9.8t + 50sqrt(3). Set to 0. t = 50sqrt(3)/9.8 = 250sqrt(3)/49. Correct. Step 3: Plug t into y(t). Correct. Step 4: Plug t into y(t) and set to 0. Correct. Step 5: Plug t into x(t). Correct. The solution seems mathematically sound. Is there a 'style' issue? The prompt asks for 'horizontal range'. The solution calculates it. Maybe the issue is that the solution doesn't explicitly state the final answers in the requested format or units? Or perhaps the 'unchecked' equations are just intermediate steps. The prompt says 'Each line... is either an EQUATION... or a SENTENCE'. The review is for the sentences. There is only one sentence: line 1. Line 1 is correct. The other lines are equations. The verdict should be pass? Let me re-read the instructions. 'Check that... the setup models the problem... each theorem... each conclusion follows... nothing would teach a student something false.' The solution is correct. However, often these adversarial reviews catch subtle sign errors or definition errors. Let's look at the range formula. Range = v_x * T. v_x = 50. T = 250sqrt(3)/49 + 8sqrt(2945)/49. Range = 12500sqrt(3)/49 + 400sqrt(2945)/49. This is correct. Is it possible the problem implies landing at the same height? No, '2 m above the ground'. Is it possible g=10? No, g=9.8. I will mark it as pass.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.