1-D Kinematics Problem/Free Fall. "A helicopter carrying Dr. Evil..." NOTE: I'm new here and I wanted to make sure I am not breaking the rules: I show most of my work step by step for this problem since there are so many things going on in it. My question is whether or not my though process is correct. I do know that one of the rules of the forum is that you should not do the problem for someone (and showing my work for the problem ends up completing a good 90% of it). I hope this is ok until I get confirmation whether or not I tackled this problem correctly. 1. The problem statement, all variables and given/known data A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s^2. Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10 seconds, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. a) What is the maximum height above the ground reached by the helicopter? b) Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0m/s^2. How far is Powers above the ground when the helicopter crashes to the ground? 2. Relevant equations a) what is the maximum height above the ground that the helicopter reaches? b) After stepping out of the helicopter, and deploying the jetpack 7 seconds after he exits, how high is Powers above the ground when the helicopter finally crashes to the ground 3. The attempt at a solution I was able to figure out part A: ( I don't know if I showed too much work, but I wanted to show that I didn't just grab the answer out of the back of my book ) - The problem states that the helicopter "takes off" therefore we can assume that the initial velocity of the helicopter is zero. - Before the engine is shut off, we know that the helicopter has a constant acceleration of 5.0 m/s^2 and that the engine was on for 10 seconds after lift off. - With an initial velocity of 0 m/s, an acceleration of 5.0m/s^2 and a time of 10s, I used the kinematics equation Δy = V0yt + 0.5(ay)t^2 to find out that the helicopter had a height of 250m when the engine was shut off. - After the engine is shut off, the chopper is still moving up, but since the acceleration is now due to gravity, it is slowing down and will eventually reach the peak where its acceleration is 0. To find how high the chopper went up AFTER the engines were shut off I made my initial velocity the velocity of the chopper when the engines were shut off: Vfy = Viy + ay*t gave me a velocity of 50m/s when the engine was shut off. - Using an initial velocity of 50m/s, an acceleration of -9.8 m/s^2 , and a final velocity of 0 (when the helicopter reaches it's highest point). The height that he helicopter traveled after the engine shut off was 127.55m. Adding the two heights together (before and after the engine shut off) gives me 377.5m above the ground (my book rounds this up to 380, I will use this value for the duration of the problem). PART b) Powers deploys a jet pack strapped on his back 7.0s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0m/s^2. How far is Powers above the ground when the helicopter crashes to the ground? Before I continue writing the rest of the attempt at a solution, I wanted to make sure my thought process is correct here If powers is stepping out of the helicopter right after the engine shuts off, is it correct to say that his initial velocity (when he jumps out) is that of the helicopter's when the engine shuts off (the 50 m/s that I calculated in the previous part) ? -If that is the case: Since powers has an initial velocity of 50m/s, I need to find out how long it takes to reach his peak (velocity of 0) after he jumps out. (And since the helicopter and Powers have the same velocity, it should take them the same time to reach the peak) - Using an initial velocity of 50 m/s , acceleration of 9.8 m/s , and a final velocity of 0. The time it takes for Powers and the helicopter to reach their peak after the engine shuts down, is 5.10 s. - Now both the helicopter and Powers are moving in the downward direction. Since he deploys the jetpack 7s after jumping out, and 5.10 seconds have already elapsed, the remaining time for (which we will be moving downward) is 1.9 seconds. - Putting Powers on the back burner for just a minute, I need to calculate how long it takes for the helicopter to hit the ground after it reaches it's peak: Using an initial height of 380m, initial velocity of 0m/s, and an acceleration of 9.8 m/s^2 I calculated a time of 8.80s. - Going back to Powers, I need to calculate the distance that he travels from when he starts falling down, to when he starts up the jetpack. As well as his velocity when he starts up the jetpack: Using an initial velocity of 0, acceleration of 9.8m/s, and a time of 1.9s (the time left until he opens up the jetpack). I get a value of 17.689m for the distance, and 18.62 m/s as the velocity when the jetpack is deployed - Since the helicopter and Powers both started falling downwards at the same time. We know that it takes 8.80s for the helicopter to hit the ground and 1.9s until the jetpack is deployed. That means that there are 6.9s left after the jetpack is deployed until the helicopter comes crashing down: Using an initial velocity of 18.62m/s, a time of 6.9s, and an acceleration of 2.0 m/s^2 (the acceleration that the jetpack provides). I got a distance of 176.088. - Adding 176.088 to the distance he traveled downwards BEFORE the jetpack gives me 193.957m. Subtracting this from the total height of 380m gives me a final answer of 186.0m. The page in my book that has the answer to this question was damaged (I was only able to read the answer to the first part) so I can't check it to be sure. Though if anything, my main concern is to get some confirmation on whether or not my though process for this problem is correct since there is so much going on.