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A body is projected vertically up with a velocity of $50 m / s$ . The percentage of its initial kinetic energy converted into potential energy after 3s is ( g = $10 m s^{- 2}$ )

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$25 %$

$64 %$

$50 %$

$87.5 %$

answer is B.

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Detailed Solution

$\frac{l o s s o f K E}{K E_{i}} \times 100 = \frac{\frac{1}{2} m (u^{2} - g^{2} t^{2})}{\frac{1}{2} m u^{2}} \times 100$

$= \frac{2500 - 900}{2500} \times 100 = 64 %$

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