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Projectile motion

Problem 9.890 · medium

A projectile is fired from \( \displaystyle 10 \) ft above the ground at \( \displaystyle 128 \) ft/s, \( \displaystyle 60^\circ \) above the horizontal. Taking \( \displaystyle g = 32 \) ft/s², find the maximum height, the time of flight and the horizontal range.
  1. x(t) = v₀cos θ · t and y(t) = 10 + v₀sin θ · t − ½gt².
    Reviewed
  2. \[ \left. \frac{d}{d t} \left(- 16 t^{2} + 64 \sqrt{3} t + 10\right) \right|_{\substack{ t=2 \sqrt{3} }} = 0 \]
    The vertical velocity is 0 at t = 2*sqrt(3).✓ Proved
  3. \[ \left. - 16 t^{2} + 64 \sqrt{3} t + 10 \right|_{\substack{ t=2 \sqrt{3} }} = 202 \]
    The maximum height.✓ Proved
  4. \[ \left. - 16 t^{2} + 64 \sqrt{3} t + 10 \right|_{\substack{ t=2 \sqrt{3} + \frac{\sqrt{202}}{4} }} = 0 \]
    It lands (y = 0) at t = 2*sqrt(3) + sqrt(202)/4.✓ Proved
  5. \[ \left. 64 t \right|_{\substack{ t=2 \sqrt{3} + \frac{\sqrt{202}}{4} }} = 128 \sqrt{3} + 16 \sqrt{202} \]
    The horizontal distance then.✓ Proved
Answer \( H = 202 \approx 202.00,\ T = 2 \sqrt{3} + \frac{\sqrt{202}}{4} \approx 7.017,\ \text{range} = 128 \sqrt{3} + 16 \sqrt{202} \approx 449.11 \)

✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. How we verify

The full receipt
LineStatusChecked byDetail
1Reviewedreviewers: gpt-oss:20b, qwen3.6:27b-mlxa sentence; read, not computed
2✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the flight stepped forward in time steps of 10⁻⁴ s

Reviewers

  • gpt-oss:20b: pass
  • qwen3.6:27b-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.
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.
  • gpt-oss:20b: pass 2026-11-06
  • qwen3.6:27b-mlx: pass 2026-11-05 — The solution correctly models the projectile motion with initial height, identifies the time of maximum height via zero vertical velocity, calculates the maximum height, solves for the landing time, and computes the horizontal range. All steps are logically sound and algebraically verified.
  • gpt-oss:20b: pass 2026-11-05
  • qwen3.6:27b-mlx: pass 2026-11-05 — The solution correctly models the projectile motion with the given initial conditions and derives the correct expressions for maximum height, time of flight, and range. The algebraic steps are consistent with the stated answer.
  • 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.