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Two paper screens $A$ and $B$ are separated by 150 m. A bullet pierces $A$ and $B$ . The hole in $B$ is 15 cm below the hole in $A$ . If the bullet is travelling horizontally at the time of hitting $A$ , then the velocity of the bullet at $A$ is: $(g = 10 m s^{- 2})$

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$100 \sqrt{3} m / s$

$200 \sqrt{3} m / s$

$300 \sqrt{3} m / s$

$500 \sqrt{3} m / s$

answer is D.

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Detailed Solution

range $= 150 = u t$ and $h = \frac{15}{100} = \frac{1}{2} \times g t^{2}$

or $t^{2} = \frac{2 \times 15}{100 \times g} = \frac{30}{1000}$ or $t = \frac{\sqrt{3}}{10}$

$∴$ $u = \frac{150}{t} = \frac{150 \times 10}{\sqrt{3}} = 500 \sqrt{3} m s^{- 1}$

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