Asked by Anonymous
A soccer player kicks a rock horizontally off the a 40 m high cliff into the pool of water. if the player hears the sound of the splash 3 sec., what is the initial speed given to the rock? Assume the speed of the sound in air to be 343 m/s.
Answers
Answered by
Damon
Time for rock to fall 40 meters?
d = (1/2) a t^2
40 = (1/2)(9.8) t^2
t^2 = 8.16
t = 2.86 seconds
so the sound traveled for 3-2.86 = .143 seconds
so the rock hit 343*.143 = 49 meters away
that is hypotenuse of angle
with leg 40 and other leg is horizontal distance d from bottom of cliff to splash
d = sqrt (49^2 -40^2)
d = 28.3 meters
so rock hit 28.3 meters from cliff base in its 2.86 meters in the air
speed = distance/time = 28.3/2.86 = 9.9 m/s
d = (1/2) a t^2
40 = (1/2)(9.8) t^2
t^2 = 8.16
t = 2.86 seconds
so the sound traveled for 3-2.86 = .143 seconds
so the rock hit 343*.143 = 49 meters away
that is hypotenuse of angle
with leg 40 and other leg is horizontal distance d from bottom of cliff to splash
d = sqrt (49^2 -40^2)
d = 28.3 meters
so rock hit 28.3 meters from cliff base in its 2.86 meters in the air
speed = distance/time = 28.3/2.86 = 9.9 m/s
Answered by
Wai
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Answered by
ZEDDO
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