Q. The kinetic energy of a projectile at the highest point is half of the initial kinetic energy. The angle of projection with the horizontal is

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A

30^o

B

45^o

C

60^o

D

90^o

Solution:

$\left(\text{KE}\right)_{\text{athiat point}} = \frac{1}{2} \left(KE\right)_{\text{initial}}$
$\frac{1}{2} m 2 \left(v \textit{cos} \text{\theta }\right)^{2} = \frac{1}{2} \left(mv\right)^{2}$
$cos^{2} \text{\theta } = \frac{1}{2}$
$\theta = 45^{\text{ο}}$

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