What Are the Launch Angles for Half the Maximum Range of a Projectile?

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SUMMARY

The discussion centers on determining the launch angles for a projectile to achieve half of its maximum range, defined by the equation R = v0² sin(2θ) / g. The key finding is that when sin(2θ) = 0.5, the corresponding angle θ can be calculated as 15 degrees. However, there are two valid angles for this scenario: 15 degrees and 75 degrees, derived from the properties of the sine function where sin(α) = 0.5 yields α = 30 degrees and its supplementary angle.

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  • Familiarity with trigonometric functions, specifically sine and arcsine
  • Knowledge of the kinematic equation for projectile range
  • Basic algebra for solving equations
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  • Study the derivation of the projectile motion range formula R = v0² sin(2θ) / g
  • Explore the concept of supplementary angles in trigonometry
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Students studying physics, particularly those focusing on mechanics and projectile motion, as well as educators seeking to clarify concepts related to angles and range in projectile dynamics.

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Homework Statement


A projectile's horizontal range on level ground is R =v0^2 sin2(theta)/g. At what launch angle or angles will the projectile land at half of its maximum possible range.


The Attempt at a Solution



So what i did was:
Half the range occurs when sin(2θ) = 0.5

So 2θ = arcsin(0.5)...so θ = 0.5*arcsin(0.5) = 15 (degrees)

but when i enter that in it says there is more than one answer, how do i get the other answer?
 
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If sin (alpha)=0.5, alpha can be 30 or 180-30 degrees.

ehild
 

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