A ball thrown up vertically returns to the thrower after 6s. Find
(a) the velocity with which it was thrown up. (b) the maximum height it reaches, and (c) its position after 4s.

Explanation

Here, time of ascent =time of descent,`t=6/2=3s`
(a) `u=?,v=0, a=-g=-9.8 m//s^(2)`
From `v=u+a t, 0=u-9.8xx3,u=29.4 m//s`
(b) From `v^(2)-u^(2)=2 as , 0-(29.4)^(2)=2(-9.8)h, h=(29.4xx29.4)/(2xx9.8)=44.1m`
(c ) At `t=3 s` , ball is at maximum height In the next 1 sec `(=4s-3s)`
From `s=ut+1/2at^(2), h=0+1/2xx9.8(1)^(2)=4.9m` (below the top)