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A bullet hits a Sand box with a velocity of $20\ m/s$ and penetrates it up to a distance of $6\ cm$. Find the deceleration of the bullet in the sand box.
Initial velocity $u=20\ m/s$
Final velocity $v=0$
Distance $s=6\ cm=\frac{6}{100}\ m$
Let $a$ be the acceleration
On using the equation of motion $v^2=u^2+2as$
$0^2=20^2+2a\times\frac{6}{100}$
Or $0=400+\frac{6a}{50}$
Or $\frac{6a}{50}=-400$
Or $6a=-20000$
Or $a=-\frac{20000}{6}$
Or $a=-3333.33\ m/s^2$
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