Question

If a stone is thrown vertically up and it is caught after time $t$ seconds, then the maximum height reached by it is ______

Answer 1

Let the initial velocity be $u$, maximum height is $h$.

Velocity at maximum height $v=0$

Time taken in upward motion=Time taken in downward motion
$=\frac{t_0}{2}$

For upward motion,

$v=u-gt$

$u=\frac{g{t}_0}{2}$

$h=ut-\frac{g{t}^2}{2}$

$h=\frac{g{t}_0}{2}\frac{t_0}{2}-\frac{1}{2}g\frac{t_0^2}{2}$

$h=\frac{g{t}_0^2}{8}$