How High Will a Failing Rocket Reach?

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Homework Help Overview

The problem involves a rocket with a mass of 7600 kg that accelerates upward at 2.35 m/s² before experiencing engine failure at a height of 570 m. The discussion centers around calculating the maximum height the rocket will reach after engine failure, the time until it returns to the launch pad, and its speed just before impact.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relevance of mass in the context of the problem and question how to set it up correctly. Some suggest using projectile motion formulas while others emphasize the need to break the problem into two parts due to non-constant acceleration.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the appropriate equations to use. There is a recognition that the acceleration changes after engine failure, prompting a division of the problem into phases of motion.

Contextual Notes

Some participants note that the rocket starts from rest and question the use of initial velocity in the equations being applied. There is also a discussion about the implications of using gravity as the only force acting on the rocket after engine failure.

QuarkCharmer
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Homework Statement


A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s^2 and feels no appreciable air resistance. When it has reached a height of 570 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


Vertical movement equations:
\Delta x=v_{i}t + \frac{at^{2}}{2}
v_{f} = v_{i} + at
v_{f}^{2} = v_{i}^{2} = 2a \Delta x


The Attempt at a Solution



I think the 7600kg is irrelevant, we have not gone over any equations that take mass into account. I'm not sure how to set this problem up.
 
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I'm using the projectile motion formula:
\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}

But since it's firing straight up, I let alpha be 90 degrees and so:

570 = 2.35t + \frac{1}{2}(-9.8)t^{2}

How can I get it's max height out of this?
 
Since the acceleration is not constant over the entire flight, you cannot use a constant acceleration equation for the whole thing. You must break the problem into two parts. Each with constant acceleration.
 
QuarkCharmer said:
I'm using the projectile motion formula:
\Delta y = v_{i}sin(\alpha)t +\frac{1}{2}(-9.8)t^{2}

But since it's firing straight up, I let alpha be 90 degrees and so:

570 = 2.35t + \frac{1}{2}(-9.8)t^{2}

How can I get it's max height out of this?

2.35 is given as the upward acceleration, it is not the initial velocity like you are trying to use it as. You are using gravity for acceleration which is not correct for the upward motion when the engines are firing.

Added: The rocket is starting from rest on the launchpad.
 
Last edited:

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