Question in the attachments Assume: the speed of sound is vs = 380.2 m/s.A boy drops a stone from rest, and it falls into a water well. If he hears the sound 5.52 seconds after he drops the stone, find d, the depth of the well in meters.Assume air friction is negligible.HINT: You will have to solve a quadratic equation.

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Assume: the speed of sound is vs = 380.2 m/s.
A boy drops a stone from rest, and it falls into a water well. If he hears the sound 5.52 seconds after he drops the stone, find d, the depth of the well in meters.
Assume air friction is negligible.
HINT: You will have to solve a quadratic equation.

Expert Solution

Step 1 To Determine , the depth of water of the well in meters.

Let us assume , that time taken for stone to travel to the surface of the water-well to be $t_{1}$ , & time taken by sound to travel back from well to the boy to be $t_{2}$ .

The stone is free-falling under influence of gravity ,so the depth can be given as : $d = u t + \frac{1}{2} g t_{1}^{2}, h e r e u = 0 \Rightarrow d = \frac{1}{2} g t_{1}^{2} . . . . (1) A l s o, i t i s g i v e n t h a t s o u n d o f s t o n e h i t t i n g t h e w a t e r i s h e a r d a f t e r 5.52 s e c o n d s . \Rightarrow t_{1} + t_{_{2}} = 5.52 s e c S p p e d o f s o u n d i s g i v e n a s, v_{s} = 380.2 m s^{- 1} \Rightarrow v_{s} = \frac{d}{t_{2}} \Rightarrow d = v_{s} \times t_{2} . . . . . (2) E q u a t i n g (1) & (2), w e c a n w r i t e t h a t \frac{1}{2} g t_{1}^{2} = v_{s} \times t_{2} S u b s t i t u t i n g t_{2} = (5.52 - t_{1}), w e g e t \frac{1}{2} g t_{1}^{2} = 380.2 m s^{- 1} \times (5.52 - t_{1}) \Rightarrow 4.9 t_{1}^{2} + 380.2 t_{1} - 2098.70 = 0 \Rightarrow t_{1} = \frac{- 380.2 \pm \sqrt{144552.04 - 4 \times 4.9 \times (- 2098.70)}}{2 \times 4.9} I g n o r i n g t h e n e g a t i v e r o o t, w e c a n w r i t e t_{1} = 5.174 s a n d t_{2} = 5.52 - 5.174 = 0.34 s S u b s t i t u t i n g i n a b o v e e q u a t i o n, w e c a n w r i t e t h a t d = 380.2 m s^{- 1} \times 0.34 s = 129.268 m \Rightarrow$

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