Question

A particle is projected at $60^{\circ}$ to the horizontal with a kinetic energy $K$. The kinetic energy at the highest point is

• A. $K$
• B. $\frac{K}{2}$
• C. $\frac{K}{4}$
• D. zero

Answer 1

Answer: (C)

Here, angle of projection, $\theta=60^{\circ} $
Let $u$ be the velocity of projection of the particle
Kinetic energy of the particle at the point of projection $O$ is
$K=\frac{1}{2}mu^{2} \ldots\left(i\right)$
where $m$ is mass of the particle
Velocity of the particle at the highest point (i.e. at maximum height) is $ucos\theta$
$\therefore $ Kinetic energy of the particle at the highest point is
$K'=\frac{1}{2}m \left(u\, cos\,\theta\right)^{2} $
$=\frac{1}{2}mu^{2} \, cos^{2}\, \theta =\frac{1}{2}mu^{2}\, cos^{2}\, 60^{\circ}$
$=\frac{1}{2}mu^{2} \left(\frac{1}{2}\right)^{2}=\frac{K}{4}$ (Using $\left(i\right))$