Projectile motion
Problem 9.677 · medium
A projectile is fired from \( \displaystyle 2 \) m above the ground at \( \displaystyle 80 \) 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.
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 40 \sqrt{3} t + 2\right) \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 200*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} }} = \frac{12098}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 40 \sqrt{3} t + 2 \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} + \frac{2 \sqrt{30245}}{49} }} = 0 \]It lands (y = 0) at t = 200*sqrt(3)/49 + 2*sqrt(30245)/49.✓ Proved
- \[ \left. 40 t \right|_{\substack{ t=\frac{200 \sqrt{3}}{49} + \frac{2 \sqrt{30245}}{49} }} = \frac{8000 \sqrt{3}}{49} + \frac{80 \sqrt{30245}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{12098}{49} \approx 246.90,\ T = \frac{200 \sqrt{3}}{49} + \frac{2 \sqrt{30245}}{49} \approx 14.168,\ \text{range} = \frac{8000 \sqrt{3}}{49} + \frac{80 \sqrt{30245}}{49} \approx 566.72 \)
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: inconclusive — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function is incorrectly defined as x(t) = 40t in line 1 (and used in line 5), omitting the cos(60°) factor. Since v₀ = 80 m/s
Every verdict on record (6)
qwen3.6:27b-mlx: inconclusive 2026-10-17 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function is incorrectly defined as x(t) = 40t in line 1 (and used in line 5), omitting the cos(60°) factor. Since v₀ = 80 m/sgpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The horizontal velocity component is incorrectly used in line 5. The problem states v₀ = 80 m/s and θ = 60°, so v_x = 80 cos(60°) = 40 m/s. However, the range calculation in line 5 substitutes t into 40*t, which is correct for x(t) if x(0)=0. Wait, let's re-read carefully. Line 1 defines x(t) = v₀cos θ · t. With v₀=80, θ=60, v₀cosθ = 40. So x(t) = 40t. Line 5 calculates x(T) = 40 * T. This is algebraically correct. Let's check the vertical motion. y(t) = 2 + 80sin(60)t - 4.9t² = 2 + 40√3 t - 4.9t². Line 2 finds t where y'=0. y' = 40√3 - 9.8t. t = 40√3 / 9.8 = 400√3 / 98 = 200√3 / 49. This matches. Line 3 calculates max height. Line 4 finds landing time. Line 5 calculates range. The solution seems correct. Why did I think it was wrong? Let me re-read the prompt's 'Stated answer'. H = 12098/49 ≈ 246.90. T ≈ 14.168. Range ≈ 566.72. Let's check the numbers. 200*sqrt(3)/49 ≈ 7.07. 2*sqrt(30245)/49 ≈ 2*173.9/49 ≈ 7.09. T ≈ 14.16. Range = 40 * 14.16 ≈ 566.4. The stated answer says 566.72. 40 * (200*sqrt(3)/49 + 2*sqrt(30245)/49) = 8000*sqrt(3)/49 + 80*sqrt(30245)/49. This matches line 5. Is there a subtle error? The problem asks for 'horizontal range'. Usually, this implies the distance from the launch point. Since x(0)=0, x(T) is the range. The setup in line 1 is standard. The equations are proved. The sentences are just labels. Is there a missing sentence explaining the method? The prompt says 'Review it'. The solution consists of 1 sentence and 4 equations. The sentence in line 1 sets up the model. The equations do the work. Is the model correct? Yes. Are the hypotheses met? Yes. Does the conclusion follow? Yes. Wait, look at line 1 again. 'x(t) = v₀cos θ · t and y(t) = 2 + v₀sin θ · t − ½gt².' This is a sentence. It is correct. The equations are marked 'proved'. I should not re-check algebra. Is there a logical gap? The solution jumps from finding t_max to finding t_land without explicitly stating 'set y(t)=0'. Line 4 says '-- It lands (y = 0) at t = ...'. This is a comment on the equation, not a separate sentence. The equation itself is `Subs(..., t, ...) = 0`. This implies the solver solved y(t)=0. This is standard. Is there any error? Let's check the value of g. g=9.8. 1/2 g = 4.9 = 49/10. The equation in line 2 is `-49*t**2/10 + 40*sqrt(3)*t + 2`. This matches y(t). The derivative is `-9.8t + 40*sqrt(3)`. Setting to 0 gives t = 40*sqrt(3)/9.8 = 200*sqrt(3)/49. Correct. The max height is y(t_max). Correct. The landing time is the positive root of y(t)=0. Correct. The range is x(t_land). Correct. Why would this fail? Maybe the 'Stated answer' is wrong? No, I review the solution against the problem. The solution matches the stated answer. Is the stated answer correct? H = 2 + (v_y^2)/(2g) = 2 + (40*sqrt(3))^2 / 19.6 = 2 + 4800/19.6 = 2 + 48000/196 = 2 + 12000/49 = (98+12000)/49 = 12098/49. Correct. T: y(t) = 0 => 4.9t^2 - 40*sqrt(3)t - 2 = 0. t = [40*sqrt(3) + sqrt(4800 + 39.2)] / 9.8. 4800 + 39.2 = 4839.2. sqrt(4839.2) = sqrt(48392/10) = sqrt(24196/5). This doesn't look like 2*sqrt(30245)/49. Let's check the term inside the sqrt in the solution's T. The solution has `2*sqrt(30245)/49`. Let's square it: 4*30245 / 2401 = 120980 / 2401 ≈ 50.38. My discriminant term: sqrt(4839.2) ≈ 69.56. 69.56 / 9.8 ≈ 7.1. 2*sqrt(30245)/49 ≈ 2*173.91/49 ≈ 7.098. They match. 30245 * 4 = 120980. 120980 / 2401 = 50.387. 4839.2 / 9.8^2 = 4839.2 / 96.04 = 50.387. Yes, it matches. The solution is correct. Why did I doubt it? Perhaps the 'sentence' in line 1 is considered insufficient? No, it's a valid setup. Is there a 'misleading' aspect? No. Is there a 'style' issue? No. It seems to pass. Wait, look at the prompt again. 'find the maximum height, the time of flight and the horizontal range.' The solution provides these. Is there any ambiguity? No. I will mark it as pass.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: pass 2026-10-14 — The solution correctly models the projectile motion with initial height, identifies the time of maximum height via zero vertical velocity, solves for the positive landing time, and computes the range. The algebraic steps are consistent with the stated answer.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.