I don't know if you've had time to think about it , but that problem had some conceptual ambiguities that most students did not seem to notice. But don't fret. Most students did the problem "correctly", even though they could have challenged the problem as not having enough information to solve. Here is the lowdown on 1(b)

Since I said that 1.8x10³ J was the energy absorbed by the detector, I should have mentioned the area of the detector. I will state a new version of the problem with two new  pieces of information: (1)The area of the detector (to be solved for )and (2) the TOTAL energy released by the fireworks explosion. I will reformulate the problem now, and if you want 7 points extra credit to go to your test score, please solve the following problem. Turn it in on Monday when you return from vacation. Note that 7 points can make the difference between an A or a B on the test

1. A fireworks rocket explodes at a height r above the ground and sends out a spherical wave. A detector D on the ground directly under the explosion measures an average sound level in decibels of 110 dB.

(a) EXTRA CREDIT ( 1 point) What is the intensity I of the sound that the detector measures ? Take the reference intensity to be I_o =1.00X10^-12 W/m². (Same solution as in-class exam.)

(b) EXTRA CREDIT ( 2 points) The TOTAL sound energy emitted by the explosion on the sphere of radius r is  1.80x10³ J. The sound signal lasts for 0.25 seconds. What is the height r above the ground ? (Same solution as in-class exam.)

(c) EXTRA CREDIT ( 4 points) The detector measures a total sound energy of 0.018 J in the same time interval of 0.25 seconds. Find the area A of the detector. (Assume the sound wave has normal incidence on the detector, which lies on and is parallel to the spherical wave front of radius r.)