Question Detail
Question
A boy throws a stone of mass 5 kg vertically downward with an initial speed of 12.0 m/s from a height 30.0 m above the ground. How long does it take the stone to reach the ground and what is the speed of the stone just before it hits the ground?
Correct Answer
Option D is the correct answer.
Detailed Explanation
Using s = ut + ½gt², with s = 30 m, u = 12 m/s, g = 9.8: 30 = 12t + 4.9t² → 4.9t² + 12t - 30 = 0. Solve: t = [-12 ± √(144 + 4×4.9×30)]/(9.8) = [-12 ± √(144 + 588)]/9.8 = [-12 ± √732]/9.8 = [-12 ± 27.06]/9.8. Positive root: (15.06)/9.8 = 1.54 s. Final velocity v = u + gt = 12 + 9.8×1.54 = 12 + 15.09 = 27.1 m/s ≈ 27.3? Actually 9.8×1.54=15.092, plus 12 = 27.092 ≈ 27.1. Option D says 1.54 s, 24.25 m/s, which is wrong. Option B says 1.53 s, 27.3 m/s. Using g=10: 30 = 12t + 5t² → 5t²+12t-30=0 → t = [-12±√(144+600)]/10 = [-12±√744]/10 = [-12±27.28]/10, positive root 15.28/10=1.528 s, v=12+10×1.528=27.28 m/s. So B is correct.
Hint
Use s = ut + ½gt² and v = u + gt.