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
$K=\frac{1}{2}m{u}^2\dots \left(i\right)$
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
$K^{\prime }=\frac{1}{2}m{\left(ucos\theta \right)}^2$
$=\frac{1}{2}m{u}^{2}co{s}^2\theta =\frac{1}{2}m{u}^{2}co{s}^{2}60^{\circ }$
$=\frac{1}{2}m{u}^2{\left(\frac{1}{2}\right)}^2=\frac{K}{4}$ (Using $\left(i\right))$