A particle is projected at an angle $60°$ with horizontal with a speed v=20 m/s. Taking $g=10 \mathrm{m}/{\mathrm{s}}^2$. Find the time after which the speed of the particle remains half of its initial speed.

$v_x=u_x=20\cos 60°=10 \mathrm{m}/\mathrm{s}$

Given, $v=\frac{u}{2}$

$$\begin{gathered} \therefore 4v^2=u^2 \\ \text{or, }4\left(v_x^2+v_y^2\right)=u^2 \\ \text{or, }4\left[(10{)}^2+v_y^2\right]={\left(20\right)}^2 \\ \therefore v_y=0 \end{gathered}$$

Hence, the projectile is at its highest point when its speed is half of its initial speed.

$\text{∴ Time taken to reach this point }t=\frac{T}{2}=\frac{u\sin \theta }{g}=\frac{\left(20\right)\sin 60°}{10}=\sqrt{3} \mathrm{s}$.