Question

A soccer player kicks a rock horizontally off a 37 m high cliff into a pool of water. If the player hears the sound of the splash 2.92 s later, what was the initial speed given to the rock (in m/s)? Assume the speed of sound in air is 343 m/s. m/s (b) What If? If the temperature near the cliff suddenly falls to 0°C, reducing the speed of sound to 331 m/s, what would the initial speed of the rock have to be (in m/s) for the soccer player to hear the sound of the splash 2.92 s after kicking the rock?

Answer 1

Answer:

(a). 16.741 m/s

(b). 15.75 m/s

Explanation:

In the question,

Height of the cliff, h = 37 m

Time taken to reach the sound to us = 2.92 s

Speed of the sound in air at room temperature = 343 m/s

Now,

Let us say the speed of the ball = u m/s

So,

Time taken by the ball to reach at the bottom, t is given by,

$t=\sqrt{\frac{2h}{g}}\\t=\sqrt{\frac{2\times 37}{9.8}}\\t=2.747\,s$

So,

Splash is heard after = 2.92 s

So,

Time taken by sound to travel the shortest distance along the hypotenuse of the triangle thus formed is,

t = 2.92 - 2.747

t = 0.172 s

Now,

Distance traveled by sound is given by,

$Distance=343\times 0.172\\Distance=59.02\,m$

So,

In the triangle using the Pythagoras Theorem,

Horizontal distance traveled is,

$D=\sqrt{59.02^{2}-37^{2}}\\D=45.99\,m$

So,

Speed of throwing of ball is given by,

$Distance=Speed\times Time\\45.99=u\times 2.747\\u=16.741\,m/s$

Therefore, the speed of the ball = 16.741 m/s.

(b).

If,

Speed of sound = 331 m/s

So,

Distance traveled by sound is,

$Distance=331\times 0.172\\Distance=56.932\,m$

So,

Distance traveled in the horizontal by ball is,

$Distance=\sqrt{56.932^{2}-37^{2}}\\Distance=43.269\,m$

So,

Speed of the ball thrown is given by,

$Speed=\frac{Distance}{Time}\\Speed=\frac{43.269}{2.747}\\Speed=15.75\,m/s$

Therefore, the speed of the ball = 15.75 m/s.