Asked by Dave
A ball is thrown directly upward. the ball is 1.35 m above the ground when thrown. The ball reaches a max height of 7.68 m above the ground. Assuming air resistance can be ignored, at what speed was the ball thrown?
Answers
Answered by
R_scott
initial KE becomes gravitational PE ... 1/2 m v^2 = m g h ... v = √(2 g h)
v = √[2 * 9.81 * (7.68 - 1.35)]
v = √[2 * 9.81 * (7.68 - 1.35)]
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