A soccer player kicks a rock horizontally off a cliff 42.9 m high into a pool of water. If the player hears the sound of the splash 3.16 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

Cliff Height, H = -42.9 m
Time to drop down cliff, t
H=0*t + (1/2)*(-g)t²
t=√(2*H/(-g))

Time for sound to travel, ts
= 3.16-t
Oblique distance,D = ts s. * 343 m/s
Horizontal distance h= √(H²+D²)

initial horizontal velocity, u
= horizontal distance / time to reach bottom
= h/t

Please check my thinking.

Answer 2

I got the right answer.
However, the equation to find the horizontal distance would be:
h = square root of (D^2 - H^2)

Answer 3

Yes, you are correct, since the oblique distance is the hypotenuse.