Question

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

Answer 1

Hint: Express the height of the building when the stone is dropped from the building using a kinematic equation. Express the kinematic equations for the time taken by the stone in both the given cases. Solve the equations you obtained.

Formula used:
Kinematic equation,
\[S = ut + \dfrac{1}{2}a{t^2}\]
Here, S is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step answer:
Let’s express the height of the building using the kinematic equation when the stone is dropped as follows,
\[h = ut + \dfrac{1}{2}g{t^2}\]
Here, g is the acceleration due to gravity.
Since the stone is dropped from the rest, the initial velocity of stone is zero. Therefore, we can write the above equation as,
\[h = \dfrac{1}{2}g{t^2}\] ……. (1)
When the stone is thrown upward, the total time taken by the stone to reach the ground is the sum of time required for the stone to go up and come back to its initial position and the time required to reach the ground from the height of the building. We assume the time taken by the stone to go up and come back to the initial position is \[{t'_1}\]. Therefore, we can write,
\[{t_1} = {t'_1} + {t_2}\] …… (2)
Here, \[{t_2}\] is the time required to reach the ground from height h.
Let’s express this time using kinematic equation as follows,
\[0 = u{{t^1}_1} - \dfrac{1}{2}g{t^1}_1^2\]
\[ \Rightarrow {t'_1} = \dfrac{{2u}}{g} = {t_1} - {t_2}\] …… (3)
Now, let’s express the time required to reach the ground when the stone is thrown downwards as,
\[h = u{t_2} + \dfrac{1}{2}gt_2^2\]
Using equation (1) in the above equation, we get,
\[\dfrac{1}{2}g{t^2} = u{t_2} + \dfrac{1}{2}gt_2^2\]
\[ \Rightarrow {t^2} = \dfrac{{2u}}{g}{t_2} + t_2^2\]

Using equation (3) in the above equation, we get,
\[{t^2} = \left( {{t_1} - {t_2}} \right){t_2} + t_2^2\]
\[ \Rightarrow {t^2} = {t_1}{t_2} - t_2^2 + t_2^2\]
\[ \Rightarrow {t^2} = {t_1}{t_2}\]
\[ \therefore t = \sqrt {{t_1}{t_2}} \]

So, the correct answer is option C.

Note: Always take the sign of acceleration due to gravity positive for the downward motion of the body and negative for upward motion. In equation (3), we have taken the distance travelled by the stone as zero since the stone comes back to the same position. The displacement in this situation is zero.