Question

David drops his lucky marble into a dry well. a) David hears the marble hit the bottom of the dry well 20 seconds later. What is the depth of the well? Assume the acceleration of gravity yo be 9.81 m/s^2 and the speed of sound to be 340 m/s. b) If David would have thrown the marble up with a speed of 5 m/s after how many seconds would have David heard the marble hit the bottom of the well?

Answer 1

total time taken = time taken to reach bottom + time taken to come up

t = sqrt(2h/g) + h/343

h is the depth of the well

20 = sqrt(2 *h/9.81) + h/343

solving for h,

h = 1292.37 is the depth of the well

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use the kinematic equation v^2 - u^2 = 2 a S

height travelled upwards

h = u^2/2g

h = 5*5/(2*9.81)

h = 1.27 m

time taken to reach top is t = sqrt(2h/g)

t = sqrt(2 * 1.27/9.81)

t = 0.5 s

total time for up and down = 0.5+ 0.5 =1

totaltime = 20+1 = 21 secs