Question

A soccer player kicks a rock horizontally off a 31.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.6 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

Answer 1

initial speed given to the rock is v= 147.46 m/s

Explanation:

Since the player kicks the rock horizontally , then its initial vertical speed is always vy₀=0 and thus the time to hit the pool is independent of the speed given to the rock . t is calculated through

0 = y + vy₀*t - 1/2*g*t² = y+ 0 - 1/2*g*t²

then

t = √(2*y/g) = √(2*31 m/ 9.8 m/s²) = 2.515 s

when the rock hits the water , the sound wave travels diagonally a distance D and is heard by the player. the distance D is

D= v sound * T

the time the sound wave travels is T= 3.6 s - 2.515 s = 1.085 s and v sound= 343 m/s

then

D= v sound * T = 343 m/s* 1.085 s = 372.155 m

since the distance D is also

D= √(x² + y²)

then the horizontal distance x is

x= √(D² - y²)= √[(372.155 m)² - (31 m)²] = 370.861 m

since the rock hits the water at t = 2.515 s , the horizontal velocity of the ball ( and thus the initial speed given to the ball since there is no acceleration in the horizontal direction if we depreciate air friction)

v=x/t = 370.861 m/2.515 s = 147.46 m/s