A particle is projected from the ground with velocity u making an angle $\mathrm{\theta }$ with the horizontal. At half of its maximum heights,

$$\begin{gathered} \begin{array}{l}{\mathrm{v}}_{\mathrm{x}}={\mathrm{u}}_{\mathrm{x}}=\mathrm{ucos}\mathrm{\theta };{\mathrm{v}}_{\mathrm{y}}=\sqrt{{\mathrm{u}}_{\mathrm{y}}^2-2\mathrm{gy}}\\ =\sqrt{(\mathrm{usin}\mathrm{\theta }{)}^2-2\mathrm{g}\frac{\left[\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2\mathrm{g}}\right]}{2}}\\ {\mathrm{v}}_{\mathrm{y}}=\sqrt{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }-\frac{{\mathrm{u}}^2{\sin }^2\mathrm{\theta }}{2}}=\frac{\mathrm{usin}\mathrm{\theta }}{\sqrt{2}}\\ \mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}=\sqrt{(\mathrm{ucos}\mathrm{\theta }{)}^2+{\left(\frac{\mathrm{usin}\mathrm{\theta }}{\sqrt{2}}\right)}^2} \\ =\mathrm{u}\sqrt{\frac{2{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }}{2}}=\mathrm{u}\sqrt{\frac{1+{\cos }^2\mathrm{\theta }}{2}}\end{array} \end{gathered}$$