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  4. A body falling from a vertical height of 10 m pierces through a distance of 1 m in sand. It faces an average retardation in sand equal to (g= acceleration due to gravity)

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Q. A body falling from a vertical height of $10\, m$ pierces through a distance of $1\, m$ in sand. It faces an average retardation in sand equal to $(g=$ acceleration due to gravity)Physics Question Image

Motion in a Straight Line

Solution:

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If the ball is dropped then $x=0$, the velocity with which it will hit the sand will be given by
$v^{2}-u^{2}=2(-g)(-9)$
$v^{2}-0=18\, g$
$v^{2}=18\, g$ ...(i)
Now on striking sand, the body penetrates into sand for $1\, m$ and comes to rest.
So, $v \rightarrow$ initial for sand and final velocity $=0$
$v^{' 2}-v^{2}=2(a) \times(-1)$
$\Rightarrow -18\, g=-2 a$
$\Rightarrow a=9\, g$