Asked by patti
If a girl drops a coin into a wishing well. After exactly 5 seconds she hears it splash into the water. How fast was the coin going after 1 second? After 2 seconds? How fast was the coin going when it hit the water? How far down is the water? Could you please help with the formulas and the answers for these questions?
Answers
Answered by
drwls
OK, here is how to do that:
How fast was the coin going after 1 second? V(1) = 1*t = 9.8 m/s
After 2 seconds? V(2) = 2*t = 19.6 m/s
How fast was the coin going when it hit the water? V(T) = 5*T
Let T be the time when it hit the water.
It will also take the sound of the splash 340*T to be heard. 340 m/s is the assumed speed of sound.
How far down is the water?
Let H be the depth of the water in the well.
sqrt(2H/g) + H/340 = 5
You will have to solve that graphically, by iteration or the quadratic formula
H = 108 meters
Time when it hit water:
T = sqrt(2H/g) = 4.69 s
V(4.69) = 46 m/s
How fast was the coin going after 1 second? V(1) = 1*t = 9.8 m/s
After 2 seconds? V(2) = 2*t = 19.6 m/s
How fast was the coin going when it hit the water? V(T) = 5*T
Let T be the time when it hit the water.
It will also take the sound of the splash 340*T to be heard. 340 m/s is the assumed speed of sound.
How far down is the water?
Let H be the depth of the water in the well.
sqrt(2H/g) + H/340 = 5
You will have to solve that graphically, by iteration or the quadratic formula
H = 108 meters
Time when it hit water:
T = sqrt(2H/g) = 4.69 s
V(4.69) = 46 m/s
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