Question

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

Answer 1

Time of ascent=time of descent $t=\frac{6}{2}=3sec$

$\left(a\right)u=?v=0a=-g=-9.8m/s^2$

$v=u+at$

$0=u-9.8\times 3$

$u=29.4m/s$

$\left(b\right)v^2-u^2=2as$

$0-{29.4}^2=2\times (-9.8)h$

$h=44.1m$

$\left(c\right)h=ut+\frac{1}{2}a{t}^2$

$h=0+\frac{1}{2}\left(9.8\right)\times 4^2=78.4m$