Question

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

Answer 1

hi,

m = 5 kg

h = 50 m

velocity = v just before hitting the ground 

 v² = u² + 2 g h = 0 + 2 * 9.8 * 50 = 980

When the stone is penetrating the sand:

v² = u² + 2 a s

0 = 980 + 2 * a * 1

a = - 490 m/s²

Average resistance Force offered by sand = m a = - 5 * 490

   =- 2, 450 N

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Answer 2

v of stone after travelling a distance of 50m =
$using - \\ {v}^{2} = {u}^{2} + 2as$
so
$so \\ {v}^{2} = {0}^{2} + 2 \times 10 \times 50 = 1000 \\ so \: v = 10 \sqrt{10}$
now after travelling 1m its velocity=0
so using
${v }^{2} = {u}^{2} + 2as$
${0}^{2} = {10 \sqrt{10} }^{2} - 2a \\ 0 = 1000 - 2a \\ so \: a = 500m \: per \: second \: square$
using
$v = u - at \\ 0 = 10 \sqrt{10} - 500t \\ so \: t = \frac{ \sqrt{10} }{50}$