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# A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at liftoff (including its fuel) is $m$, the fuel is consumed at rate $r$, and the exhaust gases are ejected with constant velocity $v_e$ (relative to the rocket). A model for the velocity of the rocket at time $t$ is given by the equation$$v(t) = -gt - v_e \ln \frac{m - rt}{m}$$where $g$ is the acceleration due to gravity and t is not too large. If $g = 9.8 m/s^2$, $m = 30,000 kg$, $r = 160 kg/s$, and $v_e = 3000 m/s$, find the height of the rocket one minute after liftoff.

## $h=14,844 \mathrm{m}$

#### Topics

Integration Techniques

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Samuel H.

University of Nottingham

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### Video Transcript

so we know that velocity is the rate of change of height. So we are going to use the fundamental theorem of calculus the second part, which tells us, um, that if we have zero to a V FTD t, this is the same thing as zero. And to go from zero to a of h prime of T gt, which is equal to And the H is just a position functions height. So we have a church of a mice h zero. We know that the initial height is zero um, so h of 00 which just gives us each of a is equal to this right here. Yeah. So this is going to give us the height of the rocket at a seconds after lunch. Um, so now, looking at our integral, what we have is the integral of a negative GT GT minus the integral of the e times. The natural log of one minus are over em times T d G. So we're going to substitute this portion right here, and we're gonna make that are you? And what we end up getting as a result is when we substitute you back after doing our integration. We're going to get a negative GT squared over to post V E M over our time is one minus are over MT mhm times a negative one plus the natural log of one minus are over em Key plus C. And this right here is going to be equal to each of a which equals the integral from zero to a a VT duty. So then what we do is we substitute in our values because we know equal 60 seconds in order to get our height after one minute. So we plug in 60 here, only evaluated, and we evaluated with our other constant values. We end up getting that. Each of 60 is approximately 14,844 m, and, um, we could do it through another method. But this is how we're going to do it. And this is our final answer, and we used it. We got to this answer using this right here, which is our height function

California Baptist University

#### Topics

Integration Techniques

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp