A steel ball is dropped from a building\'s roof and passes a window, taking .125

Question

A steel ball is dropped from a building's roof and passes a window, taking .125 second to fall from the top of the window to the bottom of the window, a distance of 1.2 meter. It then falls to a sidewalk and bounces back past the window, moving from bottom to top in .125 seconds. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the bottom of the window is 2 seconds. How tall is the building? Can you please provide all the steps. I prefer if you didnt google me an answer

Explanation / Answer

The ball reaches the window with speed V and acceleration A (9.8m/s^2) distance equation is d = 1/2At^2 + Vt (d-1/2At^2)/t = V V = (1.20m - 1/2*9.8m/s^2*0.125s^2)/0.125s V = 8.9875 m/s To get to this speed the ball fell during time T such that V = AT T = V/A = 8.9875/9.8 = 0.9171s during that time the ball fell from a height of h = 1/2At^2 = 1/2*9.8*0.9171^2 = 4.121m If the ball spends two seconds below the window, it fell for 1 second and rebounded for another 1s. In fact it reached the top of window with speed V and spent 1.125 second falling aftwerward so height below top of window is: d = 1/2At^2 + Vt d = 1/2*9.8*1.125^2 + 8.9875*1.125 = 16.3125m So total height of building is height above top of window and height below top of window h = 4.121m + 16.3125m = 20.43m

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