Question

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

Answer 1

Hint : The resistance is the retardation force of the sand on the stone. We need to make the velocity of the stone when about to hit the ground the initial velocity before collision with the ground and then calculate the answer.

Formula used: In this solution we will be using the following formula;
${v^2} = {u^2} \pm 2as$ where $v$ is the final velocity, $u$ is the initial velocity, $a$ is any acceleration including the acceleration due to gravity and $s$ is the vertical height travelled.
$v = u \pm at$ where $t$ to accelerate (or decelerate) from $u$ to $v$ .
$F = ma$ where $F$ is the force on a body, and $m$ is the mass of that body.

Complete step by step answer
A stone falls from a cliff 50 m high. The final velocity of the stone at the bottom of the cliff is the initial velocity for the travel within the sand, hence, we must calculate the velocity at the bottom of the cliff.
From ${v^2} = {u^2} + 2gs$ , initial velocity $u = 0$
By inserting all other known value we get
${v^2} = 0 + 2\left( {9.8} \right)\left( {50} \right)$
$\Rightarrow {v^2} = 980$
Taking the square root of both sides
$v = 31.3m/s$
Now, to calculate resistance, we must calculate the acceleration (actually deceleration) within the sand using the final velocity of the fall as the initial velocity of the sinking into the ground
${v^2} = {u^2} - 2as$ (negative because of deceleration, hence acting upward)
But $v = 0$ and $u = 31.3m/s$ and ${u^2} = 980$
Hence, $0 = 980 - 2a\left( 1 \right)$
$\Rightarrow 2a = 980$
Dividing both sides by 2
$a = 490m/{s^2}$
Thus, retardation is
$F = ma = 5\left( {490} \right) = 2950N$
To calculate the time for penetration, we recall
$v = u - at$
$\therefore 0 = 31.3 - 490t$
Making time subject, we have
$t = \dfrac{{31.3}}{{490}} = 0.064\operatorname{s}$ .

Note
In actuality, even if the sand has the same resistance, it wouldn’t go as deep, because air resistance would have reduced the velocity to become well less than the predicted velocity at the ground. Hence, the deepness of the stone can be taken to be the maximum depth it can ever attain being dropped from that height.