MCQ
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$g$
- ✓$9 g$
- C$100 g$
- D$1000 g$
✓
Answer
Correct option: B.
$9 g$
b
(b)
(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 $=0$
$v^{\prime 2}-v^2=2(a) \times(-1)$
$\Rightarrow -18 g=-2 a$
$\Rightarrow a=9 g$
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