Question

A stone of mass $1.0kg$ is dropped from a certain height and takes $8$ seconds for the stone to strike the sandy ground, upon reaching the ground the stone penetrates $5.0m$ into the ground. take $(g = 10m{s^{ - 2}})$ Calculate the height from where the stone is dropped?

Answer 1

Hint:In order to solve this question, we should know that when a body is dropped freely under the force of gravity its initial velocity is zero and acceleration is acceleration due to gravity so, here we will use the newton’s equation of motion to find the height from where the stone is dropped using given values of various parameters.

Formula Used:
One of useful newton’s equation of motion is,
$S = ut + \dfrac{1}{2}a{t^2}$ where,
S is the total distance covered by a body.
u is the initial velocity of a body.
a is the acceleration of a body, in case of free falling under the force of gravity $a = g = 10m{s^{ - 2}}$
t is the time at which the body covers a distance of S.

Complete step by step answer:
According to the question, we have given that
$u = 0$ stone is dropped so initial velocity is zero.
$S = H$ let H be the height from the sandy ground.
$a = g = 10m{s^{ - 2}}$ stone falling under force of gravity.
$t = 8\sec $ time taken by stone to hit the sandy ground.
Using the formula, $S = ut + \dfrac{1}{2}a{t^2}$ and putting the value of parameters we get,
$H = 0 + \dfrac{1}{2}(10){(8)^2}$
$H = 5 \times 64$
$H = 320m$
So, the distance between sandy ground and the point from where stone is dropped is $320m$ while stone also penetrate $5m$ inside the sand so, Total distance between point of dropping and the point inside the ground where stone came to rest is $325m$
Hence, height above the ground to the point of dropping the stone is $320m$ and total distance covered by the stone is $325m.$

Note:It should be remembered that, the initial velocity of a free falling body under the force of gravity is always zero and acceleration due to gravity while free fall is taken positive while if a body is thrown upwards against the force of gravity then acceleration due to gravity is taken as negative.