Rocket Engine Failure: Calculating Velocity at 1055m/s | 21.1*50 = 1055

Thread starter Jregan
Start date Sep 1, 2019
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Algebra based physics Engine Max height Physics Rocket Rocket engine

AI Thread Summary

The discussion centers on the calculation of velocity at the moment of rocket engine failure, which is determined to be 1055 m/s using the equation 21.1*50. The participants agree that this velocity marks the start of a new phase in the rocket's motion. There is uncertainty regarding the appropriate time variable for the subsequent calculations. The focus remains on understanding the implications of the engine failure on the rocket's trajectory. Overall, the conversation emphasizes the significance of accurately determining velocity in rocket dynamics.

Sep 1, 2019

Jregan

Homework Statement: A rocket accelerates vertically with a constant acceleration of 21.1m/s/s until it’s engine fails after 50 seconds. The rocket reaches an altitude of 26375m when the engine fails. When does the rocket reach its max height? (Ignoring air resistance)
I don’t understand how the equations given in the hint work with this question.

Relevant Equations: The hint says vi=a*delta t and vf=vi-g*delta t

21.1*50=1055
1055m/s is the velocity when the engine fails.
I don’t know what the time should be in the second equation.

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Sep 1, 2019

TSny

Science Advisor

Homework Helper

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Jregan said:

21.1*50=1055
1055m/s is the velocity when the engine fails.

Good. Take the instant that the engine fails to be the beginning of the second phase of the motion.

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