Question

A stone dropped from a building of height $h$ and it reaches after $t$ second on the earth. From the same building if two stones are thrown (one upwards and other downwards) with the same speed and they reach the earth surface after $t_1$ and $t_2$ seconds respectively, then
(A)$t=t_1-t_2$
(B)$t={\displaystyle \frac{t_1+t_2}{2}}$
(C) $t=\sqrt{t_1{t}_2}$
(D)$t=t_1^2{t}_2^2$

Answer 1

Hint:
When an object is in space and if there is no external force acting on the object, then the object will travel under the force of gravity. And the acceleration of the object will be equal to the acceleration due to the gravity that has a constant value of acceleration.

Complete step-by-step solution:
The height of the building is:$$h$$
The time taken by the object to cover the distance h when it release from rest is:$t$
Taking downward direction positive and upward negative.
First case:
In the first case the initial velocity of the stone is $u=0$ and time of the motion is $t$ .
Now we will use the equation of motion.
$S=ut+{\displaystyle \frac{1}{2}}g{t}^2$
Here, $S$ is the displacement and $g$ is the acceleration due to the gravity.
Substitute the values
$\begin{array}{l}h=0\times t+{\displaystyle \frac{1}{2}}g{t}^2\\ h={\displaystyle \frac{1}{2}}g{t}^2..............(1)\end{array}$
Second case: when the stone in upward direction and it reaches at time $t_1$
The initial velocity of the stone is : $u$
The displacement of the stone is: h
Now we will apply the equation of motion.
$h=-u{t}_1+{\displaystyle \frac{1}{2}}g{t}_1^2..............(2)$
Here. Negative sign is showing the direction of the velocity of the stone.
Third case: when the stone in downward direction and it reaches at time $t_2$
The initial velocity of the stone is : $u$
The displacement of the stone is: h
Now we will apply the equation of motion.
$h=u{t}_2+{\displaystyle \frac{1}{2}}g{t}_2^2$
Subtract equation (2) from equation (2)
$\begin{array}{l}0=u\left(t_1+t_2\right)+{\displaystyle \frac{1}{2}}g\left(t_2^2-t_1^2\right)\\ \Rightarrow 0=u\left(t_1+t_2\right)+{\displaystyle \frac{1}{2}}g\left(t_1+t_2\right)\left(t_2-t_1\right)\\ \Rightarrow u=-{\displaystyle \frac{g}{2}}\left(t_2-t_1\right)..............(4)\end{array}$
Now substitute the value of $h$ from the equation (1) in the equation (2).
$\begin{array}{l}\Rightarrow {\displaystyle \frac{1}{2}}g{t}^2=-u{t}_1+{\displaystyle \frac{1}{2}}g{t}_1^2\\ \Rightarrow -u{t}_1={\displaystyle \frac{1}{2}}g\left(t^2-t_1^2\right)..............(5)\end{array}$
Substitute the value of $u$ from equation (4) in the equation (5)

$\begin{array}{l}\Rightarrow -\left(-{\displaystyle \frac{g}{2}}\left(t_2-t_1\right)\right)t_1={\displaystyle \frac{1}{2}}g\left(t^2-t_1^2\right)\\ \Rightarrow t_1{t}_2-t_1^2=t^2-t_1^2\\ \Rightarrow t^2=t_1{t}_2\end{array}$
Take root both side,
$\Rightarrow t=\sqrt{t_1{t}_2}$
Therefore, the relation between $t$, $t_1$ and $t_2$ is $t=\sqrt{t_1{t}_2}$ and the correct answer is $$\mathrm{o}\mathrm{p}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\left(\mathrm{C}\right)$$ .

Note:
The relation between velocity, distance, time and acceleration can be calculated by using the equations of motion.