Projectile motion
Problem 9.897 · medium
A projectile is fired from \( \displaystyle 10 \) m above the ground at \( \displaystyle 40 \) m/s, \( \displaystyle 45^\circ \) 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) = 10 + v₀sin θ · t − ½gt².Reviewed
- \[ \left. \frac{d}{d t} \left(- \frac{49 t^{2}}{10} + 20 \sqrt{2} t + 10\right) \right|_{\substack{ t=\frac{100 \sqrt{2}}{49} }} = 0 \]The vertical velocity is 0 at t = 100*sqrt(2)/49.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 20 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{100 \sqrt{2}}{49} }} = \frac{2490}{49} \]The maximum height.✓ Proved
- \[ \left. - \frac{49 t^{2}}{10} + 20 \sqrt{2} t + 10 \right|_{\substack{ t=\frac{100 \sqrt{2}}{49} + \frac{10 \sqrt{249}}{49} }} = 0 \]It lands (y = 0) at t = 100*sqrt(2)/49 + 10*sqrt(249)/49.✓ Proved
- \[ \left. 20 \sqrt{2} t \right|_{\substack{ t=\frac{100 \sqrt{2}}{49} + \frac{10 \sqrt{249}}{49} }} = \frac{4000}{49} + \frac{200 \sqrt{498}}{49} \]The horizontal distance then.✓ Proved
Answer \( H = \frac{2490}{49} \approx 50.82,\ T = \frac{100 \sqrt{2}}{49} + \frac{10 \sqrt{249}}{49} \approx 6.107,\ \text{range} = \frac{4000}{49} + \frac{200 \sqrt{498}}{49} \approx 172.72 \)
Lines: 4 proved, 1 reviewed. 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 | Reviewed | reviewers: gpt-oss:20b, qwen3.6:27b-mlx | 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: pass — The solution correctly models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range using standard kinematic equations.
Every verdict on record (6)
qwen3.6:27b-mlx: pass 2026-11-06 — The solution correctly models the projectile motion with the given initial conditions and accurately computes the maximum height, time of flight, and horizontal range using standard kinematic equations.gpt-oss:20b: pass 2026-11-06qwen3.6:27b-mlx: pass 2026-11-05gpt-oss:20b: pass 2026-11-05qwen3.6:27b-mlx: fail (error) 2026-11-05 — The solution uses an incorrect coefficient for the gravity term in the y(t) equation (using 49/10 instead of 9.8/2 = 4.9). While 49/10 equals 4.9, the standard form is 1/2 * g, and the CAS check implies the model used was -4.9t^2, but the text in line 1 says -1/2gt^2. More critically, the time of flight calculation in line 4 solves for y=0, but the quadratic formula application seems to have a sign error or algebraic mistake in the constant term handling relative to the stated answer structure, or simply the 'unchecked' label hides that the specific radical simplification sqrt(498) is correct for the discriminant of 4.9t^2 - 20sqrt(2)t - 10 = 0? Let's check: t = (20sqrt(2) + sqrt(800 + 196)) / 9.8 = (20sqrt(2) + sqrt(996)) / 9.8. sqrt(996) = 2*sqrt(249). So t = (20sqrt(2) + 2sqrt(249))/9.8 = (10sqrt(2) + sqrt(249))/4.9 = (100sqrt(2) + 10sqrt(249))/49. The solution has 100sqrt(2)/49 + 10sqrt(249)/49. This matches. However, the first line states y(t) = 10 + v0sin(theta)t - 1/2gt^2. The CAS checks use -49/10 t^2. 49/10 = 4.9. 1/2 * 9.8 = 4.9. So the model is consistent. Wait, looking closer at line 2: Derivative of -4.9t^2 + ... is -9.8t + ... Setting to 0 gives t = 20sqrt(2)/9.8 = 200sqrt(2)/98 = 100sqrt(2)/49. This is correct. Line 3: Max height. Correct. Line 4: Time of flight. Correct. Line 5: Range. Correct. Why did I think it was wrong? Let me re-read the prompt. 'find the maximum height, the time of flight and the horizontal range'. The solution provides these. Is there a subtle error? The problem asks for 'time of flight'. Usually, this is the total time from launch to landing. The solution calculates this correctly. The 'unchecked' equations are just CAS verifications. The sentences are just the setup. Is the setup correct? Yes. Is the conclusion correct? Yes. Why would it fail? Maybe the 'unchecked' label is a trap? No, the instructions say 'do not re-check the algebra of a line marked proved'. But these are marked 'unchecked'. Does that mean I should check them? The prompt says 'Each line ... is either an EQUATION ... (its result is shown; 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 didn't check them? If I must check them, let's look at line 4 again. y(t) = 0. 4.9t^2 - 20sqrt(2)t - 10 = 0. t = [20sqrt(2) + sqrt(800 - 4(4.9)(-10))] / 9.8 = [20sqrt(2) + sqrt(800 + 196)] / 9.8 = [20sqrt(2) + sqrt(996)] / 9.8. sqrt(996) = sqrt(4*249) = 2sqrt(249). t = [20sqrt(2) + 2sqrt(249)] / 9.8 = [10sqrt(2) + sqrt(249)] / 4.9 = [100sqrt(2) + 10sqrt(249)] / 49. The solution says t = 100sqrt(2)/49 + 10sqrt(249)/49. This is correct. Line 5: x = 20sqrt(2) * t. x = 20sqrt(2) * (100sqrt(2) + 10sqrt(249))/49 = (2000*2 + 200sqrt(498))/49 = (4000 + 200sqrt(498))/49. Correct. So the math is right. Is there a style issue? The first line is a sentence. It sets up the model. It is correct. The subsequent lines are equations. The prompt asks to review 'worked calculus solutions'. The 'sentences' are what I am here for. There is only one sentence. It is correct. So why fail? Maybe the prompt implies I should find an error in the *logic* or *setup*? The setup is standard. The values are correct. Perhaps the 'unchecked' status means I need to verify the CAS output? I did, and it's correct. Is there a hidden trap? '45 degrees above the horizontal'. v0x = 40 cos 45 = 20sqrt(2). v0y = 40 sin 45 = 20sqrt(2). Correct. g=9.8. Correct. Initial height 10. Correct. Everything seems perfect. Let me look at the 'Stated answer' vs 'Solution'. The solution derives the stated answer. If the solution is correct, I should pass. But the system prompt says 'Your job is to find what is wrong'. Is there something wrong? Let's look at the time of flight again. The quadratic formula gives two roots. One positive, one negative. The solution takes the positive one. This is correct. Is it possible the question implies the projectile lands at y=10? No, 'above the ground' usually implies y=0 is ground. 'fired from 10m above the ground'. So landing is y=0. Correct. I will pass it.gpt-oss:20b: fail (error) 2026-11-05 — The time of flight is computed incorrectly: the quadratic solution gives t=10/49(10√2+√151), not 10√249/49. The height and range formulas are based on this wrong time, so the results are inconsistent with the correct physics.
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.