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 t_1 is the t
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Question 1208889: 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 t_1 is the time (measured in seconds) that it takes for the object to strike the water, then t_1 will obey the equation s = 16(t_1)^2, where s is the distance (measured in feet).
It follows that t_1 = sqrt{s}/4. Suppose that t_2 is the time that 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 t_2 to travel the distance s will be t_ 2 = s/(1100). Now t_1 + t_2 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 the water is 4 seconds. Answer by ikleyn(52800) (Show Source):
Solution
4 = +
Let y = be new variable. Then the equation takes the form
4 = + .
Multiply both sides by 1100 to get
= 0
= = .
Only positive root works y = = 15.16.
Then x = = 230 ft (rounded)
CHECK. + = 4.0 seconds. ! correct !
ANSWER. The distance to the water surface is about 230 ft.
