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)

$\frac{K}{4}$

Answer 2

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 $\left(i\right))$