SOLUTION: The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If x is the tim

Question 1207668: The distance to the surface of the water in a well can sometimes be found by dropping an object into the well and measuring the time elapsed until a sound is heard. If x is the time (measured in seconds) that it takes for the object to strike the water, than x will obey the equation s = 16x^2, where s is the distance (measured in feet). It follows that x = sqrt{s}/4.
Suppose that y is the time it takes for the sound of the impact to reach your ears. Because sound waves are known to travel at a speed of about 1100 feet per second, the time y to travel the distance s will be y = s/1100. Now x + y is the total time that elapses from the moment that the object is dropped to the moment that a sound is heard.
We have the equation TOTAL TIME ELAPSED = (sqrt{s}/4) + (s/1100).
Find the distance to the water's surface if the total time elapsed from dropping a rock to hearing it hit water is 4 seconds.

Wow! That's a mouth full.

I rearranged the original equation to be
s + 275(sqrt{s}) = 4400

Stick here....
Answer by ikleyn(52786) (Show Source): You can put this solution on YOUR website!
.


From this point, you need to solve this equation

    s +  = 4400.


Let u =   be new variable. Then your equation (1) takes the form

    u^2 + 275u = 4400,

or, equivalently,

    u^2 + 275u - 4400 = 0.


Apply the quadratic formula

     =  = .


You are looking for positive solution "u", so, you accept the positive value, only, and deny the negative one.

So, you have

     u =  = 15.16384641...


Now you need to find  s  from equation   = 15.16384641.

Square both sides and get  s = 15.16384641^2 = 229.9422379 feet.


So, the ANSWER  is : the distance to the water surface is  230 feet  (rounded).

That's all: at this point, the problem is just solved to the end.

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