A steel ball is dropped from a building\'s roof and passes a window, taking 0.12

Question

A steel ball is dropped from a building's roof and passes a window, taking 0.125 s to fall from the top to the
bottom of the window, a distance of 1.20 m. It then falls to a sidewalk and bounces back past the window, moving from
bottom to top in 0.125 s. Assume that the upward flight is an exact reverse of the fall. The time the ball spends below the
bottom of the window is 2.00 s. How tall is the building ?

Explanation / Answer

let the length of the building above top of window = l1 let the height of window = l2 let the height of the building below bottom of the window = l3 so, total height of the building l = l1+l2+l3 now velocity v at top of window = 0+gt1 = gt1 l2 = gt1*t2 + 0.5g(t2)^2 => (l2-0.5g(t2)^2)/gt2 = t1 also l1 = 0.5g(t1)^2 velocity at bottom of window = g(t1+t2) l3 = g(t1+t2)t3 + 0.5g(t3)^2 so, l = l1+l2+l3 => l = 0.5g(t1)^2 + l2 + g(t1+t2)t3 + 0.5g(t3)^2 where t1 = (l2 -0.5g(t2)^2)/gt2 => t1 = (1.2 -0.5*10* 0.125^2)/10*0.125 = 0.9s =>l = 0.5*10*0.9^2 + 1.2 + 10(0.9+0.125)1 + 0.5*10*1 ( t2 = 2/2 = 1s) = 20.5 m

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