A baseball player throws a 0.600-kg ball straight up with an initial velocity of

Question

A baseball player throws a 0.600-kg ball straight up with an initial velocity of 26.0 m/s. The ball goes straight up and comes straight back down, and the player catches it at the exact same height at which he threw it. (Ignore air resistance.) a. Find the maximum height reached by the ball. b. Find the ball's velocity at the moment when its displacement is halfway up to its maximum height. c. What is the ball's average velocity over its entire trip? d. What is the ball's average speed over its entire trip?

Explanation / Answer

a) v^2 = u^2 -2gh v=0 u=26 m/s g=9.8 m/s^2 2*9.8*h = 26^2 h = 34.49 m b) v^2 = u^2 -2g(h/2) u=26 m/s g=9.8 m/s^2 h = 34.49 m v^2 = 338 v = 18.38 m/s c) The ball's average velocity over its entire trip is 0 because velocity is vector quantity and upward downward motion will cance each other. d) The ball's average speed = total distance travelled / total time taken = 2h / 2t = h/t v = u - gt 0 = 26 - 9.8t t = 2.65s therefore The ball's average speed = 13 m/s PLEASE RATE IF IT HELPS YOU

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