Question
a man drops a stone in a well and hears the sound of the stone hitting the water after 4 sec calculate the depth of the well ( assuming the speed of sound in the well is 364 m\s )
Answers
You need the kinematic equations to solve this problem.
Also, you can either calculate this assuming that the time it takes for the sound is negligible, or you can calculate it as two parts.
If you calculate assuming the sound is instantaneous:
V(initial)=0m/s
V(final)=doesn't matter
a = 9.8m/s^2
d = ?
t = 4s
d=0*(4)+0.5*9.8*4^2
d=78.4m
If you assume that the time for the sound matters, then here is what you do:
Total time = 4s = T1 + T2
Time for object to fall
V(initial)=0m/s
V(final)=doesn't matter
a = 9.8m/s^2
d = ?
t = T1
d=0*(4)+0.5*9.8*T1^2
Next:
Time for sound
V = 364 m/s
d = d
t = 4 - T1
V*t = d
364*(4-T1) = d
1456 - 364*T1 = d
T1 = (1456 - d)/364
Now put the two together:
d=0*(4)+0.5*9.8*[(1456 - d)/364]^2
You'll probably have to use the quadratic equation to solve for d, and I doubt it will make more than a meter of difference.
Hope this helps.
Also, you can either calculate this assuming that the time it takes for the sound is negligible, or you can calculate it as two parts.
If you calculate assuming the sound is instantaneous:
V(initial)=0m/s
V(final)=doesn't matter
a = 9.8m/s^2
d = ?
t = 4s
d=0*(4)+0.5*9.8*4^2
d=78.4m
If you assume that the time for the sound matters, then here is what you do:
Total time = 4s = T1 + T2
Time for object to fall
V(initial)=0m/s
V(final)=doesn't matter
a = 9.8m/s^2
d = ?
t = T1
d=0*(4)+0.5*9.8*T1^2
Next:
Time for sound
V = 364 m/s
d = d
t = 4 - T1
V*t = d
364*(4-T1) = d
1456 - 364*T1 = d
T1 = (1456 - d)/364
Now put the two together:
d=0*(4)+0.5*9.8*[(1456 - d)/364]^2
You'll probably have to use the quadratic equation to solve for d, and I doubt it will make more than a meter of difference.
Hope this helps.