A steel ball is dropped from the roof of a building . An obsever standing in front of a window `1.5m` high notes that the ball takes `1//10s` 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 and the ground is perfectly elastic, then find the height of the building ?Take `g=10m//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`