A ball is dropped from the certain height on the surface of glass. It is collide elastically the comes back to initial position. If this process it repeated then velocity time graph is:-

Answer

(i) `100 m//s` (ii) 980 m (iii) 1600 m (iv) `(80hat(i)-140 hat(j))`

Explanation

`20=(0.6 v)/g+sqrt(2/gxx[((0.6v)^(2))/(2g)+800])` ...(i)
(i) By solving equation (i), we get `v=100 m//s`
(ii) Maximum height:
`=800+((0.6v)^(2))/(2g)=800+((0.6xx100)^(2))/20=980 m`
(iii) horizontal distance
=Horizontal velocity `xx`time of flight
`=100 cos 37^(@)xx20=1600m`
(iv) horizontal component
`v_(H)=u_(H)=100 cos 37^(@)=80 m//s`
`v_(v)=u_(v)-10xx20=100 sec 37^(@)-200`
`=140 m//s`
`:. v_("strike")=80hat(i)-140hat(j), |vec(v)|=sqrt(80^(2)+140^(2))`