A body falling from a vertical height of 10 ,m pierces through a distance of 1 ,m in sand. It faces

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2.Motion in Straight Line

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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)

A

$g$

B

$9 g$

C

$100 g$

D

$1000 g$

Solution

(b)

If the ball is dropped then $x=0$, the velocity with which it will hit the sand will be given by

$v^2-u^2=2(-g)(-9)$

$v^2-0=18 g$

$v^2=18 g$

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 $=0$

$v^{\prime 2}-v^2=2(a) \times(-1)$

$\Rightarrow -18 g=-2 a$

$\Rightarrow a=9 g$

Standard 11

Physics

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