time=time down + time up
time=sqrt(2h/g)+ h/vsound
solve for time Use the quadratic equation.
A spelunker drops a stone from rest into a hole. The speed of sound is 343m/s in air, and the sound of the stone striking the bottom is heard 1.46s after the stone is dropped. How deep is the hole? What is it in meters
4 answers
Solve the equation
t1 + t2 =
(stone's time required to fall) + (time to hear the splash) = 1.46 s
(1/2) g t1^2 = H
t1 = sqrt (2H/g)
a H = t2 (a is the sound speed, 343 m/s)
t2 = H/a
You will have to set it up as a quadratic equation in H, the depth of the well.
t1 + t2 =
(stone's time required to fall) + (time to hear the splash) = 1.46 s
(1/2) g t1^2 = H
t1 = sqrt (2H/g)
a H = t2 (a is the sound speed, 343 m/s)
t2 = H/a
You will have to set it up as a quadratic equation in H, the depth of the well.
10.24
a stone is dropped into a well . The sound of the splash is heard 30s after the stone is dropped . What is the depth of the well? Note that the speed of sound in air is 343m / s.
3 answers