A projectile with speed $50m{s}^{-1}$ is thrown at an angle of ${60}^o$ with the horizontal. The maximum height that can be reached is (acceleration due to gravity = $10m{s}^{-2}$)

Speed of projectile, $u=50m{s}^{-1}$  
Angle of projection, $\theta ={60}^o$ 
$g=10m{s}^{-2}$ 
Maximum height attained by projectile,
$H=\frac{u^2{\sin }^2\theta }{2g}=\frac{{50}^2\times {\sin }^2{60}^o}{2\times 10}=\frac{2500\times {\left(\frac{\sqrt{3}}{2}\right)}^2}{20}=125\times \frac{3}{4}=\frac{375}{4}=93.75m$