Half of maximum possible projectile range

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SUMMARY

The discussion focuses on determining the launch angles for a projectile to achieve half of its maximum horizontal range, defined by the equation R = v0²sin(2θ)/g. The maximum range occurs at a launch angle of 45 degrees, where sin(2θ) equals 1. To find the angles for half the maximum range, the value of sin(2θ) must equal 0.5, resulting in two solutions: 15 degrees and 75 degrees.

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  • Knowledge of trigonometric functions, specifically sine
  • Familiarity with the concept of maximum range in projectile motion
  • Basic algebra for solving equations
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Homework Statement



A projectile's horizontal range on level ground is R=v02sin2[tex]\vartheta[/tex]/g. At what launch angle or angles will the projectile land at half of its maximum possible range

Homework Equations





The Attempt at a Solution



All I know is that a projectile goes the furthest when the angel is 45o. I know I have to say 1/2R = the whole thing but I am not sure what to do to solve this. There are two answers. Thanks for any help! I appreciate it.
 
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Look at your formula for the range. What's special about 45° ?
 
gneill said:
Look at your formula for the range. What's special about 45° ?

It makes sin2x equal to 1. I GOT IT! I want the value where sin2x=.5 and that's at 15 and 75. Thank you!
 

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