Question

A ball whose kinetic energy is $E$, is thrown at an angle of $45$ with the horizontal, its kinetic energy at the highest point of its flight will be :

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

Answer 1

Answer: (C)

$\frac{E}{2}$

Let $\theta ={45}^o$ is the angle made by the ball with the horizontal.

At highest point there is only horizontal velocity $v_h=vcos\theta =v\cos {45}^o=\frac{v}{\sqrt{2}}$
Kinetic energy $E=\frac{m{v}^2}{2};E\propto v^2$

if velocity decrease by a factor of $\sqrt{2}$ then kinetic energy will decrease by a factor of 2.

Kinetic energy at the highest point of its flight will be $\frac{E}{2}$.