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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Updated on: 10-Oct-2022

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