How High Will a Failing Rocket Reach?

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SUMMARY

The discussion focuses on calculating the maximum height a 7600 kg rocket will reach after its engines fail at 570 m, given a constant upward acceleration of 2.35 m/s². The rocket's ascent can be divided into two phases: powered ascent and free fall. The maximum height is determined by first calculating the velocity at engine failure and then using kinematic equations to find the additional height gained during free fall. The time until the rocket impacts the launch pad and its final velocity before impact are also derived using the appropriate kinematic equations.

PREREQUISITES
  • Understanding of kinematic equations for vertical motion
  • Knowledge of projectile motion concepts
  • Familiarity with the effects of gravity on free-falling objects
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate the maximum height using kinematic equations after determining the velocity at engine failure
  • Learn how to analyze motion in two phases: powered ascent and free fall
  • Study the effects of gravitational acceleration on objects in free fall
  • Explore advanced projectile motion problems involving variable acceleration
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Physics students, aerospace engineering enthusiasts, and anyone interested in understanding the dynamics of rocket flight and projectile motion.

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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