You drop a rock into a deep well You cant see the rocks impa

You drop a rock into a deep well. You can\'t see the rock\'s impact at the bottom, but you hear it after 5 seconds. The depth of the well is  feet. Ignore air resistance. The time that passes after you drop the rock has two components: the time it takes the rock to reach the bottom of the well, and the time that it takes the sound of the impact to travel back to you. Assume the speed of sound is 1100 feet per second.

***THIS IS THE QUESTION*** ===> This is the general version of the preceding problem. Suppose the speed of sound is c feet per second, and gravity is g feet per second squared. If you drop the rock at time x=0 seconds then the depth after x seconds is d(x) = \\frac{1}{2}gx^2 feet. Suppose you hear the impact after t seconds. Then the depth of this well is d = ?

Solution

Suppose d is the depth of the well,

Let the t1 be the time for the rock to reach the bottom of the well.

And t2 the time for the sound to reach the top

Then we have

  t1+t2 = 5

depth of the well for rock travel    d = gt1² /2

from sound travel we will have d= 1100t2

equating both we have

gt1² /2 = 1100t2

eliminating t2 from the above

16t1² = 1100(5-t1)    ----------- we take g= 32ft/s²

16t1² +1100t1-5500 =0

Solving the above we get

    t1 = 4.68 s

    t2   = 0.32 s

depth of the well d = 1100*0.32 = 352 ft.

speed of sound = c

let t1 be the time for the rock to reach bottom and t2 the time for the sound to reach top

t= t1+t2

depth of the well d = gt1² /2 = ct2 = c(t-t2)

gt1² + 2ct1 -2ct =0

solving the above quadratic equation we get

t1 = (-c+sqrt(c² + 2gct))/g, we ignore the –ve root of t1

(gt1)² = c² +c² + 2gct-2c*sqrt(c² + 2gct)

depth of the well

d = gt1²/2 = {c² + gct –c*sqrt(c² + 2gct)}/g