A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sound up to 10 cm before coming to rest. Find the retardation of the ball in sand assuming it to be uniform.

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Solution

Consider the motion of ball from A toB. B $\to$ just above the sand (just to penetrate)

u = 0, a = $9.8 m / s^{2}$ S = 5 m

From $S = u t + \frac{1}{2} a t^{2}$

$\Rightarrow 5 = 0 + \frac{1}{2} (9.8) t^{2}$

$\Rightarrow t^{2} = \frac{5}{4.9} = 1.02$

$\Rightarrow t = 1.01$

$∴ v e l o c c i t y a t B, v^{'} = + a t$

= $9.8 \times 1.01 = 9.89 m / s$
....[Since u = 0]

From motion of ball in sand,

$u_{1} = 9.89 m / s, v_{1} = 0$

S = 10 cm = 0.1 m

$S_{2} = \frac{v_{1}^{2} - u_{1}^{2}}{2 a}$

= $0 - (9.89)^{2} 2 \times 0.01 = - 490 m / s^{2}$

Hence retaradation in sand is $490 m / s^{2}$

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