Question

A diving board is at a height of \( h \) from the water surface. A swimmer standing on this board throws a stone vertically upward with a velocity 16 ms\(^{-1}\). It reaches the water surface in a time of 5 s. In the next 0.2s, the diver can hear the sound from the water surface. The speed of sound is (acceleration due to gravity \( g = 10 \) ms\(^{-2} \))

• A. \( 450 \) ms\(^{-1} \)
• B. \( 225 \) ms\(^{-1} \)
• C. \( 200 \) ms\(^{-1} \)
• D. \( 275 \) ms\(^{-1} \)

Hint:

Use the equations of motion to solve problems involving constant acceleration.

Answer 1

The Correct Option is B

Let h be the height of the diving board. The stone is thrown upward with initial velocity \( u = 16 \) m/s. The acceleration is \( a = -g = -10 \) m/s\(^2\). The time taken to reach the water surface is \( t = 5 \) s.

Using the equation of motion: 

\[ s = ut + \frac{1}{2} at^2 \]

We have:

\[ -h = 16(5) + \frac{1}{2}(-10)(5^2) \]

Simplifying:

\[ -h = 80 - 125 \]

\[ -h = -45 \]

\[ h = 45 \text{ m} \]

Thus, the height of the diving board is 45 m.

The time taken for the sound to travel back to the diver is 0.2 s. Let v be the speed of sound. Then, the distance traveled by the sound is \( h = 45 \) m.

Using the formula:

\[ v = \frac{h}{t} \]

We get:

\[ v = \frac{45}{0.2} = \frac{450}{2} = 225 \text{ m/s} \]