Calculate Horizontal Range of Projectile - Unit Vector Velocity

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SUMMARY

The discussion focuses on calculating the horizontal range of a projectile given its velocity vector of (9.0i + 4.5j) m/s at 2.9 seconds after being fired. The initial attempt at a solution incorrectly applies the range formula without considering the time factor and the correct units. The accurate calculation requires integrating the time of flight and the horizontal component of the velocity to determine the range. The correct approach involves using the horizontal velocity and the time of flight to compute the range accurately.

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  • Understanding of projectile motion principles
  • Familiarity with vector components in physics
  • Knowledge of trigonometric functions, specifically arctan
  • Ability to apply kinematic equations for motion
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  • Explore the impact of time on projectile range calculations
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This discussion is beneficial for physics students, educators, and anyone interested in mastering projectile motion calculations and enhancing their problem-solving skills in kinematics.

George3
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Homework Statement


Exactly 2.9 seconds after a projectile is fired it has the velocity (9.0i +4.5j) m/s.
What is the horizontal range of the projectile.


Homework Equations





The Attempt at a Solution


Range = [((9.0i^2 + 4.5j^2)^1/2)^2 x sin2(arctan(4.5/9))]/ 9.8 m/s/s
Range = 8.3 m but this is incorrect.
 
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Of course that's not correct. Look at the units, for one thing. For another, you aren't accounting for that 2.9 seconds.
 

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