Horizontal range is maximum if the angle of projection is $45°$ and is given by $R_{\max }=\frac{u^2}{g}$

Here, $R_{max}=100 \mathrm{m}$

Then, $100=\frac{u^2}{g} \cdots \left(1\right)$

If the ball is thrown vertically upward, then the ball will attain the maximum height from the ground as $H_{\max }=\frac{u^2}{2g}$

$H_{\max }=\frac{100}{2}=50 \mathrm{m}$