1. Homework Problem 2.47 Step 0: Think! This is a kinematics problem because it has a rock falling (constant acceleration) and sound waves traveling through air (constant, and zero) acceleration. In this problem, a rock falls down a cliff, and sound waves travel back up. We are given the speed of sound in air and the total time to hear the sound after the rock is dropped. We want the cliff height. What happens? The rock falls to the bottom. Sound waves are created and the sound travels back up. This is really two separate problems, so we need to proceed in two steps: (1) rock falling and (2) sound returning. In the slides that follow, click the right mouse button or press “enter” to proceed through the solution.

2. The Falling Rock Problem 1. Draw a basic representative sketch. Just a rock falling down a cliff of height H.. 2. Draw and label vectors for relevant dynamical quantities. 3. Draw an axis, indicating origin and direction. 4. Indicate and label with appropriate subscripts the initial and final positions (include time too).

3. 5. OSE (you can “zero out” those quantities known to be zero). 6. Replace generic quantities with information given. 7. Solve algebraically. 8. Substitute numbers and box the final answer. No, not yet! We have one equation, two unknowns, H (the cliff height) and t 2 (the time it takes the rock to fall. We must find another OSE to get us a second equation. We must go to the second half of the problem.

4. The “Rising Sound” Problem 1. Draw a basic representative sketch. A “squiggle” sound wave going up. 2. Draw and label vectors for relevant dynamical quantities. 3. Draw an axis, indicating origin and direction. (Use same axis as before.) 4. Indicate and label with appropriate subscripts the initial and final positions (include time too).

5. Comments: the previous “slide” took some thinking and re-doing (that’s why we use the blackboard--erasing is easy). I already used y 1, v 1, t 1 and y 2, v 2, t 2 for the rock. I can’t re-use these symbols in the second part. Instead of using subscripts 3 and 4, I decided uppercase symbols Y, V, and T would go with the sound. IMPORTANT: the sound wave starts up at time T 1, which also the time at which the rock hit the bottom, t 2. Thus T 1 = t 2. Also, the sound reaches the top at time T 2, which is the total time, 3.4 s. HOWEVER, the time that goes in the OSE is the time it takes the sound to travel from bottom to top, or 3.4-t 2. This is the part that “catches” most people. This source of confusion will not appear on any exam problems. Comments: the previous “slide” took some thinking and re-doing (that’s why we use the blackboard--erasing is easy). I already used y 1, v 1, t 1 and y 2, v 2, t 2 for the rock. I can’t re-use these symbols in the second part. Instead of using subscripts 3 and 4, I decided uppercase symbols Y, V, and T would go with the sound. IMPORTANT: the sound wave starts up at time T 1, which also the time at which the rock hit the bottom, t 2. Thus T 1 = t 2. Also, the sound reaches the top at time T 2, which is the total time, 3.4 s. HOWEVER, the time that goes in the OSE is the time it takes the sound to travel from bottom to top, or 3.4-t 2. This is the part that “catches” most people. This source of confusion will not appear on any exam problems. Now proceed with the math…

6. 5. OSE for “rising sound” (you can “zero out” those quantities known to be zero). 6. Replace generic quantities with information given. 7. Solve algebraically. Already done! 8. Substitute numbers and box the final answer. No, not yet! For the “rising sound” part, we have one equation, two unknowns, H (the cliff height) and t 2 (the time it took the rock to fall in the first part). We must solve simultaneously the two equations from the rock falling and sound rising parts.

7. Here are our two equations: H = ½ g t 2 2 H = V (3.4 - t 2 ) There are several approaches to solving. One approach is to solve for H by eliminating t 2. To me, this is the logical thing to try (why use two steps when one will do) but the algebra became very messy. The other approach is to solve for t 2 by eliminating H, and then substituting the numerical value for t 2 back into one of the equations to get H. This proved not too difficult. Either way, because of the t 2 2, you have to solve a quadratic. The algebraic part of the solution is posted separately. Here are our two equations: H = ½ g t 2 2 H = V (3.4 - t 2 ) There are several approaches to solving. One approach is to solve for H by eliminating t 2. To me, this is the logical thing to try (why use two steps when one will do) but the algebra became very messy. Here are our two equations: H = ½ g t 2 2 H = V (3.4 - t 2 ) There are several approaches to solving. One approach is to solve for H by eliminating t 2. To me, this is the logical thing to try (why use two steps when one will do) but the algebra became very messy. The other approach is to solve for t 2 by eliminating H, and then substituting the numerical value for t 2 back into one of the equations to get H. This proved not too difficult. Either way, because of the t 2 2, you have to solve a quadratic.