A stone is released from rest from the edge of a well at time t 0 = 0 s. When it

Question

A stone is released from rest from the edge of a well at time t0 = 0 s. When it hits the bottom, a sound wave is generated. You can think of the sound wave as an object traveling from the bottom of the well towards the surface at a constant velocity vs = 340 m/s. At time t = 2.6 s, the sound of the stone reaches the top of the well. How deep is the well? At what time did the rock touch the bottom. Use 9.8 m/s2.

Hint: Use a symbol (e.g., t1) to label the time at which the sound wave is generated and starts moving.

Hint: You will get two solutions for depth. How can you check which is the correct one?

Explanation / Answer

let the depth of the well is d ,

Now , time taken by the stone to fall be t1

using second equation of motion

d = 0.5 * g * t1^2

sqrt(2d/g) = t1

Now , time taken for sound wave to travel upwards , t2 = d/vs

t2 = d/340

Now, as the total time taken is t = t1 + t2

2.6 = d/340 + sqrt(2d/9.8)

now , solving for d

d = 30.9 m

the depth of the well is 30.9 m

time taken by stone to hit ground , t1 = sqrt(2*d/g)

t1 = sqrt(2 * 30.9/9.8)

t1 = 2.511 s

the time taken by rock to hit the bottom of well is 2.511 s

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