A stone of mass $5 k g$ falls from the top of a cliff $50 m$ high and buries $1 m$ in sand. Find the average resistance offered by the sand and the time it takes to penetrate.

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Solution

Given, $M = 5 k g, h = 50 m$
$v e l o c i t y = v$

Just before hitting the ground

$v^{2} = u^{2} + 2 g h = 0 + 2 \times 9.8 \times 50 = 980$

When the stone is penetrating the sand

$v^{2} = u^{2} + 2 a s \Rightarrow 0 = 980 + 2 a (1) a = - 490 m / s^{2}$

Average resistance force offered by sand $= m a = (- 5) (490) = - 2450 N$

Time $t = \frac{v - u}{a} = \frac{0 - 10 \sqrt{1} 0}{- 500} \Rightarrow t = \frac{\sqrt{10}}{50}$

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A stone of mass 5kg falls from the top of a cliff 50m high and buries 1m in sand. Find the average resistance offered by the sand and the time it takes to penetrate.

Given, M=5kg,h=50m velocity=vJust before hitting the groundv2=u2+2gh=0+2×9.8×50=980When the stone is penetrating the sand v2=u2+2as⇒0=980+2a(1)a=−490m/s2 Average resistance force offered by sand =ma=(−5)(490)=−2450NTime t=v−ua=0−10√10−500⇒t=√1050

A stone of mass $5 k g$ falls from the top of a cliff $50 m$ high and buries $1 m$ in sand. Find the average resistance offered by the sand and the time it takes to penetrate.