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Physics
A cricket ball is hit at an angle of 30° to the horizontal with a kinetic energy E. Its kinetic energy when it reaches the highest point is
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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
KEAM
KEAM 2015
Motion in a Plane
A
$\frac{E}{2}$
B
0
C
$\frac{2E}{3}$
D
$\frac{3E}{4}$
E
E
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) , $E_{\text {total }}=E \cos ^{2} 30^{\circ}$
$\Rightarrow E_{\text {total }}=\frac{3 E}{4}$