Question

A body projected such that its kinetic energy at the top is 3/4th of its initial kinetic energy.what is the initial angle of projection of the projectile with the horizontal

Answer 1

Dear Student,

initial kinetic energy is ,
$\frac{1}{2}m{v}^2$
at the to there is no vertical motion, only horizontal motion is there. hence, kinetic energy at the top,
$\frac{1}{2}m{v}^2{\cos }^2\theta$
it is 3/4 th of the initial kinetic energy. that is,
$$\begin{gathered} \frac{1}{2}m{v}^2{\cos }^2\theta =\frac{3}{4}\left(\frac{1}{2}m{v}^2\right) \\ \cos \theta =\frac{\sqrt{3}}{2} \\ \theta ={30}^0 \end{gathered}$$

Regards,