A steel ball is dropped from the roof of a building .An observer standing in front of a window 1.25 m high notes that the ball takes 1/8 s to fall from the top to the bottom of the window .The ball reappears at the bottom of the window 2s after passing it on the way down,if the collision between the ball reappears at the bottom of the window 2s after passing it on the way down.If the collision between the ball and the ground is perfectly elastic,then find the height of the building?Take g=10 `m//s^(2)`.

Explanation

Since collision is perfectly elastic , the speed of the ball just before collision is equal to the speed of the ball just after collision . Hence time of descent is equal to the time of ascent. Therefore time taken by the ball to reach the ground from the bottom of th ewindow is 1sec.
Let u be the speed of the ball when it is at the top of the window
`rArr1.5=u1/(10)+1/2xx10xx1/(100).,(vecs=vecut+1/2vecat^(2)) rArr u=14.5m//s`
`:.`ball is dropped hence its initial speed is 0
It t be the time taken by the ball to acquire the speed of `14.5m//s` , then
`14.5=0+10xxt,(vecv=vecu+vecat)rArr t =1.45sec`
Hence total time of descent is given by
`T=1.45+1/(10)+1=2.55s`
If H be the height of the building , then `H=0+1/2"gt"^(2)`
`rArrH=1/2xx10xx(2.55)^(2)rArrH=32.5m`