The velocity will be minimum at the highest point.
So,x-co-ordinate is $\large\frac{400}{2}$ m ,i.e.,200m
Again,$y=h_{max}=\large\frac{v^2\sin^245^{\large\circ}}{2g}$
$\Rightarrow \large\frac{v^2}{4g}$
$\Rightarrow \large\frac{1}{4}$$\times 400m=100m$