A ball is dropped from a height, if it takes $0.2s$ to cross the last $6m$ before hitting the ground, find the height from which it is dropped. Take $g=10m/s^2$ (AS1)

Given,

Displacement, $s=6m$

Time, $t=0.2sec$

Apply kinematic equation of motion

$s=ut+\frac{1}{2}a{t}^2$

$6=u\left(0.2\right)+\frac{1}{2}\times 10\times {\left(0.2\right)}^2$

$u=29m/s$

$v=u+at$

$v=29+10\times 0.2=31m/s$

It falls freely from some height $h$, where initial velocity, $U=0$

Apply second kinematic equation

$V^2-U^2=2gh$

$h=\frac{V^2}{2g}=\frac{{31}^2}{2\times 10}=48.05m$

From $48.05m$ball is dropped.