Problem

While standing on a bridge 15.0 $\mathrm{m}$ abov…

04:11

Problem 62 Hard Difficulty

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 86.0 $\mathrm{m} / \mathrm{s}^{2}$ for 1.70 seconds, at which point its fuel abruptly runs out. Air resistance has no effect on its flight. What maximum altitude (above the ground) will the rocket reach?

Answer

1.214 $\mathrm{km}$

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Physics 101 Mechanics

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Video Transcript

and this problem, we have a rocket that has two phases in its flight. The first is when its engine is engaged and it starts from rest, so that initial velocity is zero meters per second. But it has a massive acceleration of 86 meters per second squared, which is about eight G's of exploration. But it only maintains this acceleration for 1.7 seconds. Now, this question is asking, um, after the engine cuts out, how high up will the rocket fly if we ignore air resistance, So as soon as the engine cuts off, you might think the acceleration become zero. Remember, as long as we're on the planet Earth, that acceleration is gonna be negative 9.8 meters per second squared while it's in the sky. And we're looking at how high up it will go so that final velocity is going to be zero meters per second. And this problem is looking for that maximum height, which is actually gonna be what we get if we add our two X values together because it's going to be traveling upwards during both phases here. So the tissue that kind of ties these two parts together is that whatever speed we end, that in phase one is what speed we start at in phase two. So those two values there are going to be the same thing. So that gives us a couple of goals, a couple of things to look for. Um, let's first start by figuring out what that final velocity is going to be. So in the first phase and phase one here, um, I have the not 80 and I'm looking for Final Velocity, which means I can use that first equation on the list that final velocity is equal to be not plus a times T. So all we need to do to get my final velocity here is multiply my acceleration and time together, which gives me an impressive 146.2 meters per second. So going pretty quickly, by the time we reach the end of the engine, and that value is both the final velocity of phase one. And like we said at the start, it's the initial velocity for phase two. Okay, so at this point, um, we know quite a bit of information, and we just need to find the X is for both phase one and phase two. And as soon as we have done that, um, we will be able to add them up and get our final answer there. So let's do phase one burst here. Um, I'm looking for the distance that it travels. My easiest equation for that one is going to be the 3rd 1 here. Um, you so x equals 1/2 final velocity plus initial velocity are multiplied by time. When I plug in my numbers here, I get X equals 1/2 of 146.2 plus zero. So just 146.2 multiplied by that time component of 1.7 seconds. And that gives me a distance of 124.27 meters as how far it travels during phase one. Um, then phase two need to find the distance. It travels there as well. We don't know the time of phase two, so that limits which of these equations we can work with? Um, the second equation here. Um, keep doing that today. The second equation here he works if we don't have time, so these squared equals V not squared. Plus two a x. When I plug in my known values, we get a final velocity squared of zero is equal to 146.2 squared um, minus 19.6 x. When I add that over, we get 19.6 x equals 146.2 squared and divide by 19.6 and put the rest in a calculator to get an X value of 1000 and 90 meters. So to finish his problem up, we take the distance from the first part 124.27 And add that to the distance in our second part of 1090. And we come up with a final answer of 1214.5 meters for how high up this rocket actually flies when it comes

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