Question

A body is projected at an angle $\theta$ to the horizontal with kinetic energy $E_k$. The potential energy of the body at the highest point of the trajectory is

• A. $E_k$
• B. $E_{k}co{s}^2\theta$
• C. $E_{k}si{n}^2\theta$
• D. $E_{k}ta{n}^2\theta$

Answer 1

The correct option is C $E_{k}si{n}^2\theta$

Let v be the velocity of projection and the angle of projection.
Kinetic energy at highest point
$=\frac{1}{2}m{v}^{2}co{s}^2\theta or{E}_{k}co{s}^2\theta$
Potential energy at highest point
$=E_k-E_{k}co{s}^2\theta =E_k(1-co{s}^2\theta )=E_{k}si{n}^2\theta$