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A stone is hurled straight upward. As it rises, it crosses point A with speed v and then crosses point B, which is 3 m above A, with speed v/2. How far above point B does the stone travel before momentarily stopping?
- 1 m
- 2 m
- 3 m
- 5 m
Correct answer: 1 m
Solution
Using v² - u² = 2as between A and B: v² - (v/2)² = 2(10)(3) gives (3/4)v² = 60, so v² = 80. At B the speed is v/2 = sqrt(20). The extra height above B = (v/2)² / (2g) = 20/20 = 1 m.
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