Angle and Range of Projectile Motion

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SUMMARY

The discussion centers on the formula for the range of projectile motion, R = v²sin(2θ)/g, demonstrating that an angle of 45 degrees maximizes distance. Given a constant velocity (v) and gravitational acceleration (g), the only variable affecting range is sin(2θ). The maximum value of sin(2θ) is 1, leading to the conclusion that 2θ must equal 90 degrees, thus θ equals 45 degrees. This establishes that launching a projectile at a 45-degree angle yields the greatest horizontal distance.

PREREQUISITES
  • Understanding of basic trigonometry, specifically sine functions
  • Familiarity with the concepts of projectile motion
  • Knowledge of gravitational acceleration (g)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the range formula for projectile motion
  • Explore the effects of varying launch angles on projectile distance
  • Learn about the impact of air resistance on projectile motion
  • Investigate real-world applications of projectile motion in sports and engineering
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Students in physics, educators teaching mechanics, and anyone interested in the principles of projectile motion and optimization of launch angles.

tharindu
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Using R=v^2sin2x/g how do you prove that the angle of 45 yields the maximum distance?
 
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for a given v (velocity) and since g(the gravity) is constant then the only variable that will affect the range (distance in the x-direction) is sin(2theta) , and the max. number can be obtained from sin(2theta) is 1 (since the range for the sin is between -1 and 1) ..

then you will have sin(2theta) = 1 >> 2theta = arcsin(1) >> 2theta= 90 >> theta = 45 ..

hopefully that was clear .. :)
 

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