A particle is fired with speed u making angle θ with the horizontal. Its potential energy at the highest point is

The vertical component of velocity is reduced to zero. The corresponding kinetic energy is converted into potential energy

$\text{ P.E. }=mgH = mg\left(\frac{u^2{\sin }^2\theta }{2g}\right)=\frac{1}{2}{\mathrm{mu}}^2{\sin }^2\mathrm{\theta }$