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The kinetic energy of a projectile at the highest point of its path is found to be $3 / 4^{th}$ of its initial kinetic energy. If the body is projected from the ground, the angle of projection is

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answer is B.

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Detailed Solution

$K . E_{top} = \frac{3}{4} (K . E_{i}); \frac{1}{2} m {(u cosθ)}^{2} = \frac{3}{4} (\frac{1}{2} {mu}^{2})$

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