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).

A steel ball is dropped from the roof of a building . An obsever stand

1.	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).
Answer» Solution :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`