Question

A small rocket is launched which starts from rest at ground level. At a height of h = 1.2 km, the rocket reaches a speed of vf = 329 m/s. At that height, the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward in this problem. A) Assuming constant acceleration, what is the rocket's acceleration, in meters per second squared, during the period from its launch until it runs out of fuel? B) After the rocket's engine turns off at a height of h = 1.2 km, it continues to move upward due to the velocity that it reached. What is the rocket's acceleration, in meters per second squared, during the period from engine shutoff until it returns to the ground? Ignore air resistance. C) Calculate the maximum height, in meters above ground level, that the rocket reaches.

          A small rocket is launched which starts from rest at ground level. At a height of h = 1.2 km, the rocket reaches a speed of vf = 329 m/s. At that height, the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward in this problem.

A) Assuming constant acceleration, what is the rocket's acceleration, in meters per second squared, during the period from its launch until it runs out of fuel?
B) After the rocket's engine turns off at a height of h = 1.2 km, it continues to move upward due to the velocity that it reached. What is the rocket's acceleration, in meters per second squared, during the period from engine shutoff until it returns to the ground? Ignore air resistance.
C) Calculate the maximum height, in meters above ground level, that the rocket reaches.

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Added by Luis Miguel A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Emily Anderson

08/10/2022

Step 1

We are given vf = 329 m/s and d = 1.2 km = 1200 m. Plugging in these values, we can solve for a: 329^2 = 0 + 2a(1200) 108241 = 2400a a = 108241/2400 a ≈ 45.1 m/s^2 Show more…

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A small rocket is launched which starts from rest at ground level. At a height of h = 1.2 km, the rocket reaches a speed of vf = 329 m/s. At that height, the rocket runs out of fuel, so there is no longer any thrust propelling it. Take the positive direction to be upward in this problem. A) Assuming constant acceleration, what is the rocket's acceleration, in meters per second squared, during the period from its launch until it runs out of fuel? B) After the rocket's engine turns off at a height of h = 1.2 km, it continues to move upward due to the velocity that it reached. What is the rocket's acceleration, in meters per second squared, during the period from engine shutoff until it returns to the ground? Ignore air resistance. C) Calculate the maximum height, in meters above ground level, that the rocket reaches.

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Transcript

00:01 Example i'm going to be looking at our kinematics equations for a motion under constant acceleration and we're going to be looking at an example of a toy rocket.

00:09 So we have a toy rocket launched from ground level and it's going to travel upwards to a height of 1 .2 kilometers or 1 ,200 meters.

00:25 At this point its rocket fuel is depleted and it is no longer being propelled.

00:32 And at this point, it has a velocity of 329 meters per second.

00:38 Okay, so we know we start from rest.

00:40 Fee initial equals zero.

00:42 My final velocity after the rocket runs out of fuel, is 329 meters per second.

00:51 And the first thing i want to do is find the rate at which this rocket was accelerating from the time it left the ground to the time it reaches the high.

01:00 Height of 1 .2 kilometers and runs out of fuel.

01:03 So the equation i'm going to use is v.

01:07 Final squared equals v initial squared plus 2a, delta d, where that's my final velocity.

01:20 My initial velocity, 2 is 2.

01:22 Acceleration is what we're looking for, and delta d is the change in distance.

01:27 So in this case, it'll be our change in height.

01:30 I've started from rest, as i've said, so this term drops out.

01:35 And i have a just equals the final squared over 2 delta d.

01:45 And in this case we go from initial height of zero to our final height of 1 .2 kilometers.

01:52 And that gives me an acceleration of a equals 45 .1 meters per second squared or about 4g.

02:07 All right.

02:08 Now the next question asks, what's the rocket's acceleration after the engine has burned out? well, the only force we have acting on it is the force due to gravity...

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