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

$$\begin{gathered} \text{velocity of the particle at half of the maximum height is } \\ {\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=\text{2as} \\ {\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=2\left(-\text{g}\right).\frac{{\text{u}}^{\text{2}}}{\text{4g}} \\ {\text{v}}^{\text{2}}-{\text{u}}^{\text{2}}=-\frac{{\text{u}}^{\text{2}}}{\text{2}} \\ {\text{v}}^{\text{2}}={\text{u}}^{\text{2}}-\frac{{\text{u}}^{\text{2}}}{2} \\ {\text{v}}^{\text{2}}=\frac{{\text{u}}^2}{2} \end{gathered}$$

$$\begin{gathered} \text{v}=\frac{\text{u}}{\sqrt{2}} \\ =\frac{50}{\sqrt{2}} \\ \left[\text{g}=10{\text{ms}}^{-\text{2}}\right] \\ =35.3 \\ \text{v}=35\text{m/s} \end{gathered}$$