A ball is projected vertically upward with a speed of  $50{\mathrm{ms}}^{-1}$. Then the speed at half of the maximum height is  $\left(\mathrm{g}=10{\mathrm{ms}}^{-2}\right)$

$\text{ Initial velocityu }=50\mathrm{m}/\mathrm{s},\text{ Acceleration due to gravity }\mathrm{g}=10{\mathrm{ms}}^{-2}$

$\text{ Maximum height }{\mathrm{H}}_{max}=\frac{{\mathrm{u}}^2}{2\mathrm{g}}\Rightarrow {\mathrm{Hou}}^2$

$\therefore \frac{{\mathrm{H}}_1}{{\mathrm{H}}_2}=\frac{{\mathrm{u}}_1^2}{{\mathrm{u}}_2^2}$

$\Rightarrow \frac{{\mathrm{H}}_1}{\frac{{\mathrm{H}}_1}{2}}=\frac{50\times 50}{{\mathrm{u}}_2^2}\Rightarrow 2{\mathrm{u}}_2^2=2500$

$\Rightarrow u_2^2=1250\Rightarrow {\mathrm{u}}_2=25\sqrt{2}$

${\mathrm{u}}_2=35.35{\mathrm{ms}}^{-1}$