Wednesday

November 26, 2014

November 26, 2014

Posted by **patti** on Monday, March 18, 2013 at 12:46pm.

- college physics -
**drwls**, Monday, March 18, 2013 at 1:15pmOK, 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

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