Question

A ball whose kinetic energy is E, is projected at an angle of ${45}^{\circ }$ to the horizontal. The kinetic energy of the ball at the highest point of its flight will be

• A. $E$
• B. $\frac{E}{\sqrt{2}}$
• C. $\frac{E}{2}$
• D. $Zero$

Answer 1

The correct option is C $\frac{E}{2}$

At highest point,
$u_x=ucos\theta u_x=\frac{u}{\sqrt{2}}\therefore KE=\frac{1}{2}m{u}_x^2=\frac{1}{2}m\frac{u^2}{2}=\frac{1}{2}\left(\frac{1}{2}m{u}^2\right)=\frac{E}{2}$