Finding magnitude and direction of a rocket

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SUMMARY

The discussion focuses on calculating the magnitude and direction of a model rocket's velocity after 4 seconds, given an initial velocity (V0) of 50 m/s at an angle of 35 degrees. The user successfully determined the magnitude of the velocity to be 42.289 m/s using the equations Vx = 50*cos(35) and Vy = 50*sin(35) - 4g. To find the direction, it is clarified that the angle should be calculated using the inverse tangent of the vertical and horizontal components of velocity, specifically tan(θ) = Vy/Vx, rather than using displacement components.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions, particularly tangent
  • Knowledge of kinematic equations for velocity and displacement
  • Basic skills in vector decomposition
NEXT STEPS
  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to apply trigonometric functions to resolve vectors
  • Explore the concept of velocity vectors and their components in physics
  • Investigate common pitfalls in interpreting physics problems involving direction and magnitude
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This discussion is beneficial for physics students, educators, and anyone involved in understanding projectile motion and vector analysis in kinematics.

CandyApples
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Homework Statement


A model rocket has been fired with V0=50m/s at 35 degrees above horizontal. Find magnitude and direction at t=4s.


Homework Equations


V = V0+at, displacement = v0t + .5at2


The Attempt at a Solution


I have successfully found the magnitude by setting Vx= 50*cos35 and Vy = 50sin35 -4g then taking the square root of each direction squared to find the magnitude is equal to 42.289 m/s. I am having trouble now, figuring out what to do to figure out the direction. I have calculated displacement of x to be 163.83m and displacement of y to be 36.32m by using the displacement equation. Any hints on how to find the angle? I thought inverse tangent of y/x would give me the correct displacement, however it did not.
 
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Be careful. "Find the magnitude and direction" is ambiguous, but based on the fact you got the correct answer for the magnitude by taking the magnitude of the velocity, I assume the question is "Find the magnitude and direction of the velocity", not "Find the magnitude of the velocity and the direction of the displacement", which is what you solved.
 
If you draw a triangle you will see that Vx and Vy form a right triangle with V. You can use trig(tan) to find the angle.
 

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