# Question #40610

Jan 21, 2017

As referred to the figure above.

Kinetic energy at point of projection $K {E}_{\text{projection}} = \frac{1}{2} m {u}^{2}$
Maximum height of the object having mass $m$ will be reached when $\sin \theta$ component of projectile's velocity becomes $0$ (zero).

As such kinetic energy at this point is given by
$K {E}_{\text{max height}} = \frac{1}{2} m {\left(u \cos \theta\right)}^{2}$
Ratio of the two is $\left(K {E}_{\text{max height")/(KE_"projection}}\right) = \frac{\frac{1}{2} m {\left(u \cos \theta\right)}^{2}}{\frac{1}{2} m {u}^{2}}$
$= {\cos}^{2} \theta$