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A bullet of mass $10 g m$ is fired horizontally with a velocity of $1000 m s^{- 1}$ from a height of $50 m$ above the ground. If the bullet reaches the ground with a velocity of $500 m s^{- 1},$ the work done against air resistance in Joules is $(g = 10 m s^{- 2})$

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$17.5 J$

$3750 J$

$5005 J$

$3755 J$

answer is B.

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Detailed Solution

The velocity with which a horizontal projectile reaches the ground $= \sqrt{u^{2} + 2 g h}$

Work done against air resistance is $\frac{1}{2} m (u^{2} + 2 g h) - \frac{1}{2} m v^{2}$

= $\frac{1}{2} \times 10 \times 10^{- 3} (1000^{2} + 2 \times 10 \times 50 - {(500)}^{2})$

$W = 3755 J$

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