A soccer player kicks a rock horizontally off a 40.0 m high cliff into a pool of water. If the player hears the sound of the splash 3.04 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.

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No

9.99 To find the initial speed given to the rock, we need to use the equation of motion that relates the height of the cliff, the time it takes for the sound to travel, and the speed of sound in air.

First, let's calculate the time it took for the rock to fall from the cliff to the water. Since we are dealing with vertical motion, we can use the equation:

h = (1/2)gt^2

where h is the height, g is the acceleration due to gravity (9.8 m/s^2), and t is the time. Rearranging the equation to solve for t:

t = sqrt(2h / g)

Plugging in the values:

t = sqrt(2 * 40.0 m / 9.8 m/s^2)
t ≈ 2.02 s

Next, we need to find the distance between the cliff and the pool, assuming the sound traveled at a constant speed of 343 m/s. We can use the equation:

Distance = Speed × Time

where the distance is the unknown and the speed is the speed of sound (343 m/s), and the time is the time it took for the sound to reach the player (3.04 s).

Distance = 343 m/s × 3.04 s
Distance ≈ 1044.32 m

Now, to find the initial speed given to the rock, we can use the equation that relates distance, time, and velocity (assuming the horizontal velocity is constant):

Distance = Velocity × Time

where the distance is the calculated distance between the cliff and the pool (1044.32 m), the time is the time it took for the rock to fall (2.02 s), and the velocity is the initial speed given to the rock (which is what we're trying to find).

1044.32 m = Velocity × 2.02 s

Rearranging the equation to solve for velocity:

Velocity = Distance / Time

Velocity = 1044.32 m / 2.02 s
Velocity ≈ 517.35 m/s

Therefore, the initial speed given to the rock was approximately 517.35 m/s.