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Q.
A particle is projected at $60^°$ to the horizontal with an energy $E$. The kinetic energy and potential energy at thehighest point are
Work, Energy and Power
Solution:
Initial KE, $E=\frac{1}{2}mu^{2}\,...\left(i\right)$
At the highest point, velocity $\upsilon=u\,cos\,60^{°}=\frac{u}{2}$
KE at highest point $=\frac{1}{2}m\upsilon^{2}=\frac{1}{2}m\left(\frac{u}{2}\right)^{2}$
$=\frac{1}{4}\left(\frac{1}{2}mu^{2}\right)=\frac{E}{4}$ (Using (i))
PE at highest point $=mg\times\frac{u^{2}\,sin^{2}\,60^{°}}{2g}$
$=\frac{1}{2}mu^{2}\left(\frac{\sqrt{3}}{2}\right)^{2}=\frac{3}{4}E$