Question

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

• A. $K$
• B. $zero$
• C. $K/4$
• D. $K/2$

Answer 1

Answer: (C)

$K/4$

Given: Angle of projection= ${60}^0$

Solution: Horizontal component of velocity is given by,

$v\cos {60}^0$

Therefore, kinetic energy at highest point is given by,

$K_H=\frac{1}{2}m{v}^2{\cos }^2\theta$

or

$K_H=K{\cos }^2{60}^0$

As, $\cos {60}^0=\frac{1}{2}$

So we get,

$K_H=\frac{K}{4}$

Hence, the correct option is (C).