Find the speed of the particle after 0.2 seconds

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SUMMARY

The particle is thrown horizontally with an initial speed of 10 m/s. After 0.2 seconds, the vertical velocity (vy) is calculated using the equation vy = gt, resulting in a value of 2 m/s when using g = 10 m/s². The total speed (v) after 0.2 seconds is determined using the formula v = sqrt(vx² + vy²), yielding a final speed of approximately 10.198 m/s. The discussion highlights the importance of using the correct gravitational constant, with 9.8 m/s² being the standard value for more precise calculations.

PREREQUISITES
  • Understanding of projectile motion
  • Familiarity with basic kinematic equations
  • Knowledge of gravitational acceleration (g)
  • Ability to perform vector addition of velocities
NEXT STEPS
  • Study the effects of air resistance on projectile motion
  • Learn about the differences between using g = 9.8 m/s² and g = 10 m/s² in calculations
  • Explore advanced kinematic equations for non-horizontal launches
  • Investigate the concept of trajectory and its applications in physics
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Students studying physics, educators teaching projectile motion, and anyone interested in understanding the dynamics of thrown objects.

annalian
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Homework Statement


The particle is thrown horizontally with speed 10 m/S (I believe from a height). Find the speed of it after 0.2 seconds.

Homework Equations


vx=v
vy=gt
v=sqrt(vx^2+vy^2)[/B]

The Attempt at a Solution


vx=10
vy=10*0.2=2
v=sqrt(100+4)=10.198m/S Am i right?
 
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Looks right to me. I think it is probably more common to use 9.8 ms-2 for gravity though.
 

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