A ball is dropped from rest from the top of a cliff thatis 41 mhigh. From ground

Question

A ball is dropped from rest from the top of a cliff thatis 41 mhigh. From ground level, a second ball is thrown straight upward atthe same instant that the first ball is dropped. The initial speedof the second ball is exactly the same as that with which the firstball eventually hits the ground. In the absence of air resistance,the motions of the balls are just the reverse of each other.Determine how far below the top of the cliff the balls crosspaths.

I keep getting the answer of 10.25 meters below the top of thecliff. Can someone verify this for me? Thanks!

Explanation / Answer

10.25 appears correct v t - 1/2 g t2 + 1/2 gt2  = 41   v t = 41 Where v is the initial speed of the ball thrown upward v2 = 2 g s    v = 28.35 m /s    t = 1.446 sec s = 1/2 g t2 = 10.25

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