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Suppose you were to drop a stone down a well. A falling object obeys the equation s=16(t1)2 , where s is the distance fallen in feet and t is the time in seconds. Sound waves travel at a speed of about 1100 ft/sec, and the time it takes to travel a distance s is t2=s1100 . The total time it takes for you to hear the stone hit the water below equals the time it spent falling plus the time it takes the sound to reach your ear. What is the distance to the water's surface in the well if you hear the stone hit the water 4 seconds after dropping the stone?

Sagot :

Oladayoademosuidn

Answer:

228.64 ft

Step-by-step explanation:

Given that the total time, t it takes for you to hear the stone hit the water below equals the time it spent falling, t1 plus the time it takes the sound to reach your ear t2.

Given that t = 4s and t = t1 + t2.

So t1 + t2 = 4  (1)

Also s = 16(t1)² and t2 = s/1100

Thus t2 = 16(t1)²/1000

From (1), t2 = 4 - t1

So 4 - t1 = 0.016(t1)²

0.016(t1)² + t1 - 4 = 0

(t1)² + 62.5t1 - 250 = 0

Using the quadratic formula to find t1,

[tex]t1 = \frac{-1 +/- \sqrt{1^{2} - 4 X -4 X 0.016} }{2 X 0.016}\\t1 = \frac{-1 +/- \sqrt{1 + 0.256} }{0.032} \\t1 = \frac{-1 +/- \sqrt{1.256} }{0.032} \\t1 = \frac{-1 +/- {1.121} }{0.032} \\t1 = \frac{-1 + 1.121 }{0.032} or t1 = \frac{-1 - 1.121 }{0.032}\\t1 = 0.121/0.032 or -2.121/0.032\\t1 = 3.78 s or -66.28 s[/tex]

Since t1 cannot be negative, t1 = 3.78 s

So, s = 16(t1)²

= 16(3.78)²

= 16(14.29)

= 228.64 ft

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