Question

A cricket ball is hit at an angle of ${30}^{\circ }$ to the horizontal with a kinetic energy $E$. Its kinetic energy when it reaches the highest point is

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

Answer 1

Answer: (D)

$\frac{3E}{4}$

Given, $\theta ={30}^{\circ }$
$KE=E$. We know that,
$E=\frac{1}{2}m{v}^2....\left(i\right)$
$E=\frac{1}{2}{m}^2{\cos }^2\theta$
$=\frac{1}{2}m{v}^2\cdot {\cos }^2{30}^{\circ }$
From Eq. (i), $E_{total}=E{\cos }^2{30}^{\circ }$
$\Rightarrow E_{total}=\frac{3E}{4}$.