Question

A rocket, starting from rest, travels straight up with an acceleration of 57.7 m/s². When the rocket is at a height of 520 m, it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity I. Later, the station measures an intensity 1/3I. Assuming that the speed of sound is 343 m/s, find the time that has elapsed between the two measurements.

Answer 1

The intensity is proportional to the surface area of the sphere over which the sound wave spreads, so an intensity of 1/3 the original intensity means that the distance has increased to sqrt(3) times its original value. That is, the second sound is emitted when the rocket is at a height of
520 m * sqrt(3) = 900.666 m

The rocket's motion can be described by
y = (1/2)(57.7 m/s^2)t^2.
Substituting 520 m and 900.66 m for y, we get times of
t1 = 4.24 seconds and t2 = 5.58 seconds.

The time when the first sound reaches the ground is
t(a) = 4.24 seconds + (520m)/(343 m/s) = 5.756 s
The time when the 2nd sound reaches the ground is
t(b) =5.58 seconds + (900.66 m)/(343 m/s) = 8.205 s

The time elapsed between the two ground-based measurements is 2.449 seconds