A ball is thrown vertically upwards with a velocity u from the ground. The ball attains a maximum height `H_("max")` . Then find out the time and displacement at which ball have half of the maximum speed

Explanation

Maximum speed of the ball is u. At point B and C ball have speed u/2 but direction is opposite so from

Let `t_1` is the time taken by the ball from point A to Band `t_2` is the time taken by the ball from A to C
From A to B `u/2=u-gt_1`
from A to C `-u/2=u-g t^(2)`
From (i) `t_1=u/(2g)` from (ii) `t_2=(3u)/(2g)`
For equation `v^2-u^2=2as`
`impliesv=+-u//2,u=u,a=-g`
`implies(u/2)^2-u^2=-2gh`
`h=(3u^2)/(8g){(u^2)/(2g)=H_("max"))`
`h=3/4H_("max")`