Question

A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? (Given g=10$m{{s}^{-2}}$ and speed of sound = 340$m{{s}^{-1}}$)
A. 10s
B. 11.47s
C. 1.10s
D. 20s

Answer 1

Hint: First, calculate the time taken for the stone to hit the surface of the water in the pond by using the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$. Then calculate the time taken for the sound of the splash to reach the top of the tower by using $\text{time = }\dfrac{\text{distance}}{\text{speed}}$.

Formula used:
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$\text{time = }\dfrac{\text{distance}}{\text{speed}}$

Complete answer:
It is given that a stone is dropped from the top of a tower that is 500m high. Due to the gravitational force exerted by earth on the stone, the stone will accelerate. The direction of the gravitational force is in the downward direction. Hence, the stone will accelerate downwards with an acceleration g, i.e. acceleration due to gravity.

For small heights like 500m, the value of g is constant and it is given to be equal to 10$m{{s}^{-2}}$.
Therefore, the stone will go down with this acceleration and after some time it will hit the surface of the water in the pond below. Once, the stone hits the surface, a sound will be produced. This sound will travel with the speed of sound and after some time it will reach the top.

Let us first calculate the time taken for the stone to hit the surface of the water. Let this time be ${{t}_{1}}$. To find the value of ${{t}_{1}}$ let us use the kinematic equation $s=ut+\dfrac{1}{2}a{{t}^{2}}$….. (i),

where u is the initial velocity of the stone, s is the displacement of the stone, a is the acceleration of the stone and t is the time taken for this displacement.
Let us consider the positive y axis as positive direction and the negative y axis negative direction.
In this case, u=0, s=-500m, a=-g=-10$m{{s}^{-2}}$ and t=${{t}_{1}}$.
Substitute the values in equation (i).

$\Rightarrow -500=(0){{t}_{1}}+\dfrac{1}{2}(-10)t_{1}^{2}$
$\Rightarrow 500=5t_{1}^{2}$
$\Rightarrow 500=5t_{1}^{2}$
$\Rightarrow t_{1}^{2}=100$
$\Rightarrow {{t}_{1}}=\pm 10s$

Since time cannot be negative, ${{t}_{1}}=-10s$ is discarded.
Therefore, ${{t}_{1}}=10s$.

This means that the stone will hit the surface of water after 10 seconds.
Now, let us calculate the time for the sound of the splash to reach the top. Let this time be ${{t}_{2}}$.
To find ${{t}_{2}}$ let us use the formula $\text{time = }\dfrac{\text{distance}}{\text{speed}}$.
The speed of the sound is given as 340$m{{s}^{-1}}$ and the distance it has to cover is 500m.
Therefore, ${{t}_{2}}=\dfrac{500}{340}=1.47s$.
This means that sound will reach the at the top in 1.47 seconds after the stone hits the surface.
Therefore, the time after which the sound of the splash is heard, once the stone is dropped is equal to ${{t}_{1}}+{{t}_{2}}=10+1.47=11.47s$.

So, the correct answer is “Option B”.

Note:
Some students may make a mistake by calculating only the time taken for the stone to hit the water surface (i.e. ${{t}_{1}}$) and write it as the time after which the splash is heard at the top.
The sound of the splash travels in all directions with the same speed. Different observers standing at different positions will hear the sound at different times.