Asked by jamie
Consider that a coin is dropped into a wishing well. You want to determine the depth of the well from the time t between releasing the coin and hearing it hit the bottom. Suppose that t=2.059 s and assume the speed of sound in air is 330 m/s. What is the depth of the well?
Can someone please help, I cant get the right answer.
What I did was (0.5)(-9.80)(2.059^2)= 20.77m
But it is not correct and when I incorporate 330 m/s it gives me really big value.
Can someone please help, I cant get the right answer.
What I did was (0.5)(-9.80)(2.059^2)= 20.77m
But it is not correct and when I incorporate 330 m/s it gives me really big value.
Answers
Answered by
bobpursley
You did it incorrectly.
The coin hits the water, then the sound comes up. You are given the total time.
Let t be total time, t1 be the time to fall, so t-t1 is the time for the sound to come up.
h=1/2*g*t1
h=vsound*(t-t1)
set these equal, and solve for t1, then go back and solve for h.
The coin hits the water, then the sound comes up. You are given the total time.
Let t be total time, t1 be the time to fall, so t-t1 is the time for the sound to come up.
h=1/2*g*t1
h=vsound*(t-t1)
set these equal, and solve for t1, then go back and solve for h.