A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be 10 m/s, then the maximum height attained by the stone is (Take g =10 $m/s^2$)

The correct option is D 10 m
Let $u$ be the initial velocity and $h$ be the maximum height attained by the stone.
$v_1^2=u^2-2gh$,
$(10{)}^2=u^2-2\times 10\times \frac{h}{2}$
$100=u^2-10h$ ....(i)
Again at height $h$,
$v_2^2=u^2-2gh$
$(0{)}^2=u^2-2\times 10\times h$
$u^2=20h$ ... (ii)
So, from Eqs. (i) and (ii) we have
$100=10h$
$h=10m$