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A man drops a rock into a well. (a) the man hears the sound of the splash 2.40 s after he releases the rock from rest. the speed of sound in air (at the ambient temperature) is 336 m/s. how far below the top of the well is the surface of the water? (b) what if? if the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated?

Respuesta :

Part (a)
Let h =  depth of water surface from the top of the well, m

Because the stone is dropped, its initial velocity is zero.
Let t₁ =  time for the stone to drop, s
Then
h = (1/2)*(9.8 m/s²)*(t₁ s)²
t₁ = √(h/4.9) s = 0.4518√h s

The velocity of sound is 336 m/s.
The time for the sound wave to travel from the water surface to the top of the well is
t₂ = h/336 m/s 

Because the time before the sound is heard is 2.4 s, therefore
t₁ + t₂ = 2.4
0.45181√h + h/336 = 2.4
Multiply through by 336.
151.8082√h + h = 806.4
h - 806.4 = -151.8082√h
h² - 1612.8h + 6.5028 x 10⁵ = 2.3046 x 10⁴h
h² - 2.4659 x 10⁴h + 6.5028 x 10⁵ = 0

Solve with the quadratic formula.
h = 0.5[2.4659 x 10⁴ +/- 2.4606 x 10⁴]
   = 24632 m or 26.5 m

Test the two answers.
When h = 24632,
   t₁ = 0.4518√(24632) = 70.9 s (Not acceptable)
When h = 26.5 m,
   t₁ = 0.4518√(26.5) = 2.3258 s, t₂ = 26.5/336 = 0.0789 s,
   t₁+t₂ = 2.4 s (CORRECT)

Answer:  h = 26.5 m

Part (b)
If travel time for the sound is ignored, then
0.4518√h = 2.4
h = (2.4/0.45118)² = 28.2 m
The percent error is 100*[(28.2 - 26.5)/26.5] = 6.42%

Answer: The error is  6.4%

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