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A soccer player kicks a rock horizontally off a 41.7 m high cliff into a pool of water. If the player hears the sound of the splash 3.11 s later, what was the initial speed given to the rock? Assume the speed of sound in air to be 343 m/s.
To solve this problem, we can use the equation of motion in the vertical direction:
h = (1/2) * g * t²
Where:
h = Height of the cliff (41.7 m)
g = Acceleration due to gravity (9.8 m/s²)
t = Time of flight (3.11 s)
First, we need to calculate the time it takes for the rock to fall from the cliff:
h = (1/2) * g * t²
41.7 = (1/2) * 9.8 * t²
t² = (2 * 41.7) / 9.8
t² = 8.5163265306
t ≈ √8.5163265306
t ≈ 2.92 s
Since the player hears the sound of the splash 3.11 s later, it means the rock was in flight for 3.11 - 2.92 = 0.19 s before hitting the water.
Now, let's calculate the horizontal distance traveled by the rock in 0.19 s. We can use the equation:
d = v * t
Where:
d = Horizontal distance traveled
v = Initial velocity
t = Time
Since we want to find the initial speed, we'll assume there is no acceleration in the horizontal direction.
d = v * t
d = v * 0.19
Substituting the values we know:
d = v * 0.19
To find the initial speed, we need to find the horizontal distance traveled by the rock. We can use the speed of sound in air (343 m/s) as the horizontal distance because the player heard the splash at the same time the rock hit the water.
d = 343 * 0.19
d ≈ 65.17 m
Now, we have the horizontal distance traveled. We can use this value to find the initial speed:
d = v * 0.19
65.17 = v * 0.19
v ≈ 342.47 m/s
Therefore, the initial speed given to the rock was approximately 342.47 m/s.