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Q. 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

KEAMKEAM 2015Motion in a Plane

Solution:

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