Let $u=$ velocity of projection of the ball. The ball will cover maximum horizontal distance when angle of projection with horizontal, $\theta=45^{\circ}$. Then $R_{\max }=u^2 / g$
Now, $100=u^2 / g$
In order to study the motion of the ball along vertical direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of $Y$ axis. Taking motion of the ball along vertical upward direction, we get, $u_y=u, a_y=-g, v_y=0$,
we know, $v_y=u_y+a_y t$
$\Rightarrow 0=u+(-g) t$ or, $t=u / g$
Again, $y=u_y t+1 / 2 a_y t^2$
$$
\therefore \quad y=u(u / g)+1 / 2(-g) u^2 / g^2=u^2 / g-1 / 2 u^2 / g
$$
$$
=u^2 / g=100 / 2=50 \mathrm{~m}
$$