If someone would work this problem out it would be much appreciated. I and havin

Question

If someone would work this problem out it would be much appreciated. I and having difficulty understanding what they mean by 4.6s after it hits the ground. I would appreciate.

A ball is dropped from the roof of a building. It hits the ground and it is caught at its original height 4.6 s later. (Assume the ball bounces and returns to the height from which it was dropped.)
(a) What was the speed of the ball just before it hits the ground?


(b) How tall was the building?


You are watching from a window 2.9 m above the ground. The window opening is 1.3 m from the top to the bottom. (Assume the bottom of the window is 2.9 m above the ground.)
(c) At what time after the ball was dropped did you first see the ball in the window?

Explanation / Answer

time of flight T = 4.6 s time of desent t = T / 2 = 2.3 s let the height of the building be h in downward motion: ------------------------- initial velocity u = 0 accleration a = 9.8 m / s^ 2 time t= 2.3 s from the relation v= u+at = 0+22.54 = 22.54 m / s (a)the speed of the ball just before it hits the ground v = 22.54 m / s (b) from the relation v^ 2- u^ 2= 2aS 22.54^2 -0^ 2 = 2(9.8)(h ) from this required height h = 25.921 m (c). length of the window h ' = 1.3 m height of the bottom of the window above the ground H = 2.9 m let the required time be t " distance travel by ball from top of the building to top of the window S = h -( H + h ' ) =21.721 m from the relation s = ut" + ( 1/ 2) at"^ 2 21.721 = 0 + 4.9 t"^ 2 from this t " =2.105 s

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