Question

A body is projected at angle of ${30}^o$ to the horizontal with kinetic energy $E$.The kinetic energy at the top most point is

• A. $\frac{3E}{4}$
• B. $\frac{E}{4}$
• C. $0$
• D. $\frac{E}{2}$

Answer 1

The correct option is A $\frac{3E}{4}$

Given that,

Angle $\theta ={30}^0$

Now, v is the velocity of projection

The energy is

$E=\frac{1}{2}m{v}^2$

At top the velocity

$u=v\cos {30}^0$

$u=v\times \frac{\sqrt{3}}{2}$

Now, kinetic energy at top

$K.E=\frac{1}{2}m{\left[\frac{\sqrt{3}}{2}m{v}^2\right]}^2$

$K.E=\frac{3}{4}\left(\frac{1}{2}m{v}^2\right)$

$K.E=\frac{3E}{4}$

Hence, the kinetic energy is $\frac{3E}{4}$ at top