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

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SUMMARY

The discussion centers on the calculation of velocity at the point of rocket engine failure, specifically at 1055 m/s, derived from the equation 21.1 * 50 = 1055. This velocity marks the transition to the second phase of motion following engine failure. Participants emphasize the importance of defining the time variable in subsequent calculations to accurately assess the motion dynamics post-failure.

PREREQUISITES
  • Understanding of basic physics concepts related to motion and velocity
  • Familiarity with equations of motion in kinematics
  • Knowledge of rocket propulsion and engine failure dynamics
  • Ability to interpret and manipulate mathematical equations
NEXT STEPS
  • Research kinematic equations for motion analysis after engine failure
  • Explore the principles of rocket propulsion and failure analysis
  • Learn about the impact of velocity on rocket trajectory
  • Investigate methods for calculating time variables in motion equations
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Aerospace engineers, physics students, and professionals involved in rocket design and analysis will benefit from this discussion.

Jregan
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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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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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