Projectile motion
Problem 9.893 · medium
A projectile is fired from \( \displaystyle 0 \) m above the ground at \( \displaystyle 20 \) 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) = 0 + v₀sin θ · t − ½gt².
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 10 \sqrt{3} t\right) \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} }} = 0 \]The vertical velocity is 0 at t = 50*sqrt(3)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{3} t \right|_{\substack{ t=\frac{50 \sqrt{3}}{49} }} = \frac{750}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 10 \sqrt{3} t \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} }} = 0 \]It lands (y = 0) at t = 100*sqrt(3)/49.✓ Proved
- \[ \left. 10 t \right|_{\substack{ t=\frac{100 \sqrt{3}}{49} }} = \frac{1000 \sqrt{3}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{750}{49} \approx 15.31,\ T = \frac{100 \sqrt{3}}{49} \approx 3.535,\ \text{range} = \frac{1000 \sqrt{3}}{49} \approx 35.35 \)
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 at which the vertical velocity becomes zero is t=100√3/49, not 50√3/49. Consequently the maximum height computed in step 3 is incorrect. The subsequent time of flight and range are correct, but the earlier steps contain a computational error.qwen3.6:27b-mlx: fail (error) — The horizontal position function used in step 5 is x(t) = 10t, which corresponds to a horizontal velocity of 10 m/s. However, with v₀ = 20 m/s and θ = 60°, the horizontal velocity is v₀cos(60°) = 10 m/s, so this part is actually correct. Wait, let me re-read carefully. v₀=20, cos(60)=0.5, so v_x = 10. Step 1 says x(t) = v₀cos θ · t. Step 5 uses Subs(10*t, ...). This is consistent. Let's check the vertical part. v₀sin(60) = 20 * sqrt(3)/2 = 10*sqrt(3). Step 1 says y(t) = v₀sin θ · t - 1/2 g t². Step 2 uses -49*t**2/10 + 10*sqrt(3)*t. 1/2 * 9.8 = 4.9 = 49/10. So the equation is correct. The derivative is -9.8t + 10*sqrt(3). Setting to 0 gives t = 10*sqrt(3)/9.8 = 100*sqrt(3)/98 = 50*sqrt(3)/49. This matches step 2. The max height calculation in step 3 is correct. The landing time in step 4 is 2 * t_max = 100*sqrt(3)/49. This matches. The range in step 5 is 10 * 100*sqrt(3)/49 = 1000*sqrt(3)/49. This matches. The stated answer matches the solution. Why did I think it was wrong? Let me re-read the prompt's instruction. "find what is wrong". Is there a subtle error? Step 1 defines the model. Steps 2-5 are equations marked proved. The sentences are just the setup in step 1. Step 1 is correct. The solution seems perfectly sound. Let me double check the arithmetic. H = (v₀sinθ)² / 2g = (10√3)² / 19.6 = 300 / 19.6 = 3000 / 196 = 750 / 49. Correct. T = 2v₀sinθ / g = 20√3 / 9.8 = 200√3 / 98 = 100√3 / 49. Correct. R = v₀²sin(2θ) / g = 400 * sin(120) / 9.8 = 400 * (√3/2) / 9.8 = 200√3 / 9.8 = 2000√3 / 98 = 1000√3 / 49. Correct. The solution is correct. I will mark it as pass.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-11-06 — The horizontal position function used in step 5 is x(t) = 10t, which corresponds to a horizontal velocity of 10 m/s. However, with v₀ = 20 m/s and θ = 60°, the horizontal velocity is v₀cos(60°) = 10 m/s, so this part is actually correct. Wait, let me re-read carefully. v₀=20, cos(60)=0.5, so v_x = 10. Step 1 says x(t) = v₀cos θ · t. Step 5 uses Subs(10*t, ...). This is consistent. Let's check the vertical part. v₀sin(60) = 20 * sqrt(3)/2 = 10*sqrt(3). Step 1 says y(t) = v₀sin θ · t - 1/2 g t². Step 2 uses -49*t**2/10 + 10*sqrt(3)*t. 1/2 * 9.8 = 4.9 = 49/10. So the equation is correct. The derivative is -9.8t + 10*sqrt(3). Setting to 0 gives t = 10*sqrt(3)/9.8 = 100*sqrt(3)/98 = 50*sqrt(3)/49. This matches step 2. The max height calculation in step 3 is correct. The landing time in step 4 is 2 * t_max = 100*sqrt(3)/49. This matches. The range in step 5 is 10 * 100*sqrt(3)/49 = 1000*sqrt(3)/49. This matches. The stated answer matches the solution. Why did I think it was wrong? Let me re-read the prompt's instruction. "find what is wrong". Is there a subtle error? Step 1 defines the model. Steps 2-5 are equations marked proved. The sentences are just the setup in step 1. Step 1 is correct. The solution seems perfectly sound. Let me double check the arithmetic. H = (v₀sinθ)² / 2g = (10√3)² / 19.6 = 300 / 19.6 = 3000 / 196 = 750 / 49. Correct. T = 2v₀sinθ / g = 20√3 / 9.8 = 200√3 / 98 = 100√3 / 49. Correct. R = v₀²sin(2θ) / g = 400 * sin(120) / 9.8 = 400 * (√3/2) / 9.8 = 200√3 / 9.8 = 2000√3 / 98 = 1000√3 / 49. Correct. The solution is correct. I will mark it as pass.gpt-oss:20b: fail (error) 2026-11-06 — The time at which the vertical velocity becomes zero is t=100√3/49, not 50√3/49. Consequently the maximum height computed in step 3 is incorrect. The subsequent time of flight and range are correct, but the earlier steps contain a computational error.qwen3.6:27b-mlx: inconclusive 2026-11-05 — reviewer response could not be parsed: {"verdict": "fail", "severity": "error", "notes": "The horizontal position function x(t) is incorrectly defined as 10*t in line 5, whereas v₀cos(60°) = 20 * 0.5 = 10 m/s is correct, but the solution fgpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: fail (style) 2026-11-05 — [domain objection, downgraded to style] The horizontal position function in line 1 is defined as x(t) = v₀cos θ · t, but line 5 substitutes x(t) = 10t. Since v₀ = 20 and cos(60°) = 0.5, the coefficient should be 10, which matches, BUT line 5 calculates the range using t = 100*sqrt(3)/49. Let's check the time of flight. y(t) = 20*sin(60)*t - 4.9*t^2 = 10*sqrt(3)*t - 4.9*t^2. Roots are t=0 and 10*sqrt(3)/4.9 = 100*sqrt(3)/49. This is correct. Range = 10 * (100*sqrt(3)/49) = 1000*sqrt(3)/49. This is also correct. Wait, let me re-read line 1. x(t) = v0 cos theta * t. v0=20, theta=60. cos(60)=1/2. So x(t) = 10t. Line 5 uses 10t. This is consistent. Let me check the max height. vy = 10*sqrt(3) - 9.8t. vy=0 => t = 10*sqrt(3)/9.8 = 100*sqrt(3)/98 = 50*sqrt(3)/49. Line 2 says t = 50*sqrt(3)/49. Correct. y(t) = 10*sqrt(3)*t - 4.9*t^2. At t = 50*sqrt(3)/49: y = 10*sqrt(3)*(50*sqrt(3)/49) - 4.9*(50*sqrt(3)/49)^2 = (1500/49) - 4.9*(2500*3/2401) = 1500/49 - (4.9*7500)/2401. 4.9 = 49/10. So (49/10 * 7500)/2401 = (49 * 750)/2401 = 750/49. So y = 1500/49 - 750/49 = 750/49. Correct. Why did I think it was wrong? Let me look closer at the prompt's provided solution text vs the equations. Line 1 defines y(t) = 0 + v0 sin theta t - 1/2 g t^2. Line 2 uses -49*t^2/10 + 10*sqrt(3)*t. 1/2 g = 4.9 = 49/10. v0 sin theta = 20 * sqrt(3)/2 = 10*sqrt(3). This matches. Line 5 uses Subs(10*t, ...). This implies x(t) = 10t. Line 1 says x(t) = v0 cos theta t. 20 * 1/2 = 10. This matches. Is there a subtle error? "A projectile is fired from 0 m above the ground". The setup seems correct. The calculations seem correct. The stated answer matches the equations. Is there a style issue? Line 1 is a sentence defining the model. It is correct. The equations are marked unchecked but the prompt says "do not re-check the algebra of a line marked proved". Wait, the prompt says "Each line ... is either an EQUATION ... (its result is shown; do not re-check the algebra of a line marked proved)". The lines 2-5 are marked `[equation, unchecked]`. This might mean the CAS didn't check them, or I shouldn't check them? The instructions say "do not re-check the algebra of a line marked proved". These are marked `unchecked`. This implies I SHOULD check them? Or does it mean the CAS result is not shown? The prompt says "its result is shown" for equations. Lines 2-5 show results. But they are marked `unchecked`. This is ambiguous. However, usually `unchecked` in these prompts means the CAS did not verify it, so I must verify it. I verified it and it is correct. Is there a conceptual error? "find the maximum height, the time of flight and the horizontal range." The solution finds them. The values are correct. Let me look really closely at Line 1. `x(t) = v₀cos θ · t and y(t) = 0 + v₀sin θ · t − ½gt².` This is standard. Line 2: `Subs(Derivative(-49*t**2/10 + 10*sqrt(3)*t, t), t, 50*sqrt(3)/49) = 0`. The derivative is `-49*2*t/10 + 10*sqrt(3)`. At `t = 50*sqrt(3)/49`, `-98/10 * 50*sqrt(3)/49 + 10*sqrt(3) = -9.8 * 50*sqrt(3)/49 + 10*sqrt(3) = -0.2 * 50*sqrt(3) + 10*sqrt(3) = -10*sqrt(3) + 10*sqrt(3) = 0`. Correct. Line 3: Value is 750/49. Correct. Line 4: Value is 0. Correct. Line 5: Value is 1000*sqrt(3)/49. Correct. Why would this fail? Maybe the unit? No units requested in output format, just numbers. Maybe the significant figures? The problem gives 20, 60, 9.8. 9.8 has 2 sig figs. The answer is given as exact fractions and approximations. This is standard. Is there a missing negative sign? No. Is the angle measured from the vertical? "60 above the horizontal". No. Is g positive or negative? y(t) has -1/2 g t^2. Correct. I see no error. I will pass it.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.