A ball is dropped from a height of $5 m$ onto a sandy floor and penetrates the sand up to $10 c m$ 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 $A$ to $B$ .
$B \to$ just above the sand ( just to penetrate)
$u = 0, a = 9.8 m / s^{2}, s = 5 m$

The distance covered by the ball is:
$S = u t + \frac{1}{2} a t^{2}$
$\Rightarrow 5 = 0 + \frac{1}{2} (9.8) t^{2}$

$\Rightarrow t^{2} = \frac{5}{4.9} = 1.02$
$\Rightarrow t = 1.01$ .

$∴$ velocity at $B, v = u + a t = 9.8 \times 1.01 (u = 0) = 9.89 m / s$ .

From motion of ball in sand
$u_{1} = 9.89 m / s, v_{1} = 0, a = ?, s = 10 c m = 0.1 m$

$a = \frac{v_{1}^{2} - u_{1}^{2}}{2 s} = \frac{0 - (9.89)^{2}}{2 \times 0.1} = - 490 m / s^{2}$

The retardation in sand is $490 m / s^{2}$ .

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

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