Initial K.E, K.E. =21mu2 Speed at heighest point $\begin{aligned}
& \mathrm{V}=\mathrm{u} \cos 60^{\circ}=\frac{\mathrm{u}}{2} \
& \therefore \mathrm{KE}_2=\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{u}}{2}\right)^2 \
& =\frac{1}{4} \times \frac{1}{2} \mathrm{mu}^2 \
& =\frac{\mathrm{KE}}{4}
\end{aligned}$