Question

From the top of a multi- storeyed building, $39.2\,m$ tall, a boy projects a stone vertically upwards with an initial velocity of $9.8\,m{s^{ - 1}}$ such that it finally drops to the ground. (i) When will the stone reach the ground? (ii) When will it pass through the point of projection? (iii) What will it pass through the point of projection? (iii) What will be its velocity before striking the ground? Take $g = 9.8\,m{s^{ - 2}}$.

Answer 1

Hint:Use the equations of motion formula and calculate the height of projection from the building and the time taken by the stone to reach the ground. Divide the time taken to reach the ground by two, to find the time taken to reach the point of projection. Use the equation of motion again to find the final velocity of the stone.

Useful formula:
(1) The equation of the motion is given as

${v^2} = {u^2} - 2gs$

Where $v$ is the final velocity, $u$ is the initial velocity, $s$ is the distance covered and $g$ is the acceleration due to gravity.

(2) The other equation of motion is given as

$V = u + gt$

Where $V$ is the final velocity of stone.

(3) The third equation of motion is given as

$S = ut + \dfrac{1}{2}g{t^2}$

Complete step by step solution:
It is given that the
Height of the multi- storeyed building, $h = 39.2m$
Initial velocity of the stone, $u = 9.8\,m{s^{ - 1}}$
(i) Using the equation of motion,

${v^2} = {u^2} - 2gs$

Substitute the value of the given initial velocity, final velocity as zero and the acceleration due to gravity in above equation.

${0^2} = {9.8^2} - 2 \times 9.8s$

By simplifying the above equation, we get

$s = 4.9\,m$

The height of the stone from the ground $4.9 + 39.2 = 44.1\,m$

$V = u + gt$
$V = 0 + 9.8 \times 1$

By simplification of the above step,

$V = 9.8\,m{s^{ - 1}}$

Using the third equation of motion,

$S = ut + \dfrac{1}{2}g{t^2}$

Substituting all the parameters we get

$39.2 = 9.8t + 0.5 \times 9.8{t^2}$

By doing various mathematical operations, we get

$\left( {t + 4} \right)\left( {t - 2} \right) = 0$

The time taken is taken as $2$ , since it is positive. And hence the total time to reach the ground is $2 + 2 = 4\,s$ .

(ii) The time taken to pass the point of projection is obtained by dividing the total time by

$2$ , $\dfrac{4}{2} = 2\,s$ .

(iii) Using the first equation of motion,

${V^2} = {U^2} + 2gs$

Substituting the known parameters in it,

${V^2} = {9.8^2} + 2 \times 9.8 \times 39.2$

By performing the basic arithmetic operations, we get

$V = \sqrt {96.04 + 768.32} $

By simplifying further, we get

$V = 29.4\,m{s^{ - 1}}$ .

Hence the velocity of the stone before striking the ground is $29.4\,m{s^{ - 1}}$ .

Note:At first equation of the motion, the final velocity is substituted as zero. This is because the stone reaches the at most point at that time the velocity of stone decreases to zero. While falling down, the stone moves with the acceleration due to gravity.