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.

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

Step-by-step 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