THE QUESTION IS NOT PROPERLY WRITTEN
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 t1 is the time (in seconds) it takes for the object to strike the water, then t1 obeys the equation s = 16t1², where s is the distance (in feet) Solving for t1, we have t1 = √s/4. Let t2 be the time it takes for the sound of the impact to reach your ears. Since sound waves travel at a speed of approximately 1100 feet per second, the time to travel the distance s is t2 = s/1100. Now t1 +t2 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 elapsed time √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.
Answer:
229.94 feet
Step-by-step explanation:
Given
t1 = √s/4
t2 = s/1100
From the question, we understand that
t1 + t2 = 4 seconds
Let y = √s
So,
t1 + t2 = 4 becomes
√s/4 + s/1100 = 4
y/4 + y²/1100 = 4 ------ Solve the fractions
(275y + y²)/1100 = 4
275y + y² = 4 * 1100
275y + y² = 4400 ------- Rearrange
y² + 275y - 4400 = 0 (Quadratic Equation)
The standard form of a quadratic equation is ax² + bx + c = 0
Where x =
frac{-b+-\sqrt{bx^{2} - 4ac } }{2a}
Here a = 1, b = 275 and c = -4400
So,
y = (-275+- √(275² - 4*1*-4400))/2
y = (-275+- √75625 + 17600))/2
y = (-275+- √(93225))/2
y = (-275 +- 305.328)/2
y = (-275 + 305.328)/2 or (-275-305.328)/2
y = 30.328/2 or -580.328/2
y = 15.164 or -290.164 -------Negative is not applicable
So, y = 15.164
But y = √s
So, s = y²
S = 15.164²
s = 229.94 ft
