Question

The kinetic energy of a projectile at the highest point of its path is found to be $3/4^{th}$ of its initial kinetic energy. If the body is projected from the ground, the angle of projection must be:

• A. $0^0$
• B. ${30}^0$
• C. ${60}^0$
• D. ${40}^0$

Answer 1

The correct option is B ${30}^0$

Let A be the initial point of projection and B be the point of maximum height

$K.E_B=\frac{1}{2}m{v}^2{\cos }^2\theta$

$K.E_A=\frac{1}{2}m{v}^2$

Now, $K.E_A=\frac{3}{4}K.E_B$

$\Rightarrow \cos \theta =\frac{\sqrt{3}}{2}$

$\Rightarrow \theta ={30}^{\circ }$