Question

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

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

Answer 1

Answer: (C)

The motion of projectile is as shown in figure,

Let initial velocity of projection is $u$,
so initial kinetic energy is
$K{E}_1=\frac{1}{2}m{u}^2\dots$ (i)
At highest point, the vertical component of velocity is zero,
while the horizontal component is $u\cos \theta$, which is non-zero.
So, the kinetic energy at this point is
$K{E}_2=\frac{1}{2}m(u\cos \theta {)}^2=\frac{1}{2}m{u}^2{\cos }^2\theta$
$=K{E}_1{\cos }^{2}60^{\circ }=\frac{K}{4}$
$(\because K{E}_1=K$ and $\cos 60^{\circ }=1/2)$