A ball proected vetically upwards from (A), the top fo tower reaches the ground in ` t_1` second. If it is projected vectically downwards from (A) with the same vecoty, it reaches the ground in ` t_2 secons . If it falls freely from (A), show that it would reahc the ground in ` sqrt t_1 t-2 seconds`.

Explanation

Let (h) be the height of a tower and (u) the initial velocity of projection of the ball.
(i) When ball is projected vertically upwards. Taking vertically downwards motin of ball form top of tower to ground we have ` u=- u , a g , S=h, t=t_1`
As, ` S =ut_2 + 1/2 at^2 ` :. h =- ut_1 + 1/2 xx g xx t_1^2` (i)
(ii) When ball is projected vertically downwards. Taking vertically downward motion of ball from top of tower to ground , we have ` u=u, a=g , S= h t=t_2`
:. ` h = at_2 + 1/2 gt_2^2` ...(ii)
(iii) When ball falls freely, ` u= 0 , a=g S= h , t = ?` h =1/2 gt^2` (iii)`
Multiplying (i) by ` t_1` and ` by ` t_1` and then adding , we get ` h (t_1 + t_2) = 1/2 gt_1 t_2 (t-1 + t_2) or h = 1/2 gt-1 t_2`
Putting this value in (ii) we get. ` 1/2/ gt_1 t_2 = 1/2 gt^2 ` or ` t^2 =t_1 `or ` t= sqrt_1 t_2`.