What Is the Maximum Height a Rocket Will Reach After Engine Failure?

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Homework Help Overview

The discussion revolves around a physics problem involving a rocket that experiences engine failure after reaching a certain height. The subject area includes kinematics and dynamics, specifically focusing on the motion of the rocket under the influence of gravity after the engines stop functioning.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore how to calculate the maximum height the rocket will reach after engine failure, questioning the use of various kinematic equations. There are attempts to find the final velocity at the moment of engine failure and how to use it to determine the additional height gained. Some participants express confusion about the variables involved and the relationships between them.

Discussion Status

The discussion is ongoing with multiple interpretations being explored. Participants have provided guidance on using kinematic equations to find the final velocity and subsequent height, but there is no explicit consensus on the correct approach or calculations. Some participants have shared their results, but discrepancies in answers indicate that further clarification is needed.

Contextual Notes

There is uncertainty regarding the initial velocity of the rocket at the moment the engines fail, as well as the correct application of kinematic equations. Some participants mention rounding errors in their calculations, and the problem constraints regarding the rocket's motion under gravity are under discussion.

starchild75
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Homework Statement



A 7400 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.15 m/s^2 and feels no appreciable air resistance. When it has reached a height of 560 m, its engines suddenly fail so that the only force acting on it is now gravity. a. What is the maximum height this rocket will reach above the launch pad? b. How much time after engine failure will elapse before the rocket comes crashing down to the launch pad? c. How fast will it be moving just before it crashes?

Homework Equations



560=.5(2.15)t^2.
v0=2.15t
ht=560Tv0t-.5(2.15)t^2

The Attempt at a Solution



I got 1115 meters and it's wrong.
 
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starchild75 said:
560=.5(2.15)t^2.
v0=2.15t
OK, but not necessary. You have to find the velo vf at that height. Use the other formula.

ht=560Tv0t-.5(2.15)t^2
??

Use vf as initial velo at that height and find max height. The other answers will follow.
 
How do I find t?
 
how do I find final velocity when just given acceleration and height?
 
Let me use some symbols. The rocket is at a height h when the engines stop. The velo is vf at that point. It rises to H before it falls back. I'm denoting the height of the rocket at any time by y.

1. You can find H-h if you know vf and g. Use formula for final, initial velos and height.

2. From h, you can find t for y to become zero, that is, time to fall. Use directly y = ut-1/2 gt^2. Decide on what you'll put as y.
 
starchild75 said:
how do I find final velocity when just given acceleration and height?

Vf^2 = Vi^2 + 2ah. This is when the rocket is rising with const accn 'a', OK?
 
Last edited:
What does u represent?
 
The initial velo, that is, vi. Have you understood the rest?
 
No. I have been looking at this for five hours now. I don't know the velocity or the time. So I don't know how I can solve the problem.
,
 
  • #10
does final velocity equal 2 ah?
 
  • #11
starchild75 said:
does final velocity equal 2 ah?

What does post #6 say?
 
  • #12
final velocity^2 +initial velocity^2 +2ah.

I don't know the final velocity or the initial velocity.
 
  • #13
All these formulas have two unknowns.
 
  • #14
I know when the rocket hits its maximum height, the velocity is zero.
 
  • #15
starchild75 said:
final velocity^2 +initial velocity^2 +2ah.

I don't know the final velocity or the initial velocity.

final velocity^2 =initial velocity^2 +2ah.

vi=0, a=2.15, h=560. What more do you want to find vf?

(Nothing has been said about the initial velo of the rocket, so it is assumed to be zero, if that's what was stopping you. Anyway, a rocket is not thrown upward, it rises from rest due to accn.)
 
  • #16
The final velocity is 49 m/s Now what do I do?
 
  • #17
I got 881 for height.
 
  • #18
I got 650 and it says there is a rounding error.
 
  • #19
So 650 is another attempt at finding the height?? You're going to have to show more detail of your calculations for us to spot where you might have gone wrong.
 
  • #20
starchild75 said:
The final velocity is 49 m/s Now what do I do?

Correct.

starchild75 said:
I got 881 for height.

starchild75 said:
I got 650 and it says there is a rounding error.

Decide on one, and show your calc. Incidentally, neither is correct.

Try to understand. At the point where the engines switch off, the rocket is moving under g only. It's initial velo is 49 m/s upward. So, to what height does it rise from that point? Final velo will be zero, and that's how you find the height. So, you know total height H.

For finding the time to fall to ground, use the height h=560 m as your starting posn. Use y=vi*t-1/2 gt^2. Then what will be the 'y' for the ground? What will be vi?

(Gather your thoughts and send fewer posts, instead of writing a line and posting it.)
 

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