Angle of Projection from Height and Range

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SUMMARY

The problem involves determining the angle of projection for a projectile whose horizontal range is three times its maximum height. The relevant equations are R = (Vo² sin(2θ))/g for range and H = (Vo sin(θ)²)/(2g) for height. By setting R equal to 3H, the equation simplifies to 4/3 tan(θ) = 1, leading to the conclusion that tan(θ) = 3/4. This indicates the angle of projection can be derived from the tangent function based on the established relationship between range and height.

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  • Familiarity with algebraic manipulation of equations
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Homework Statement


A projectile is fired in such a way that that its horizontal range is three times its maximum height. What is the angle of projection?


Homework Equations


R = Vo2sin(2theta)/g

H = (Vosin(theta)2/2g

R = 3H

Cancel Voo, g and sin(theta) to leave 4/3 tan(theta) = 1


The Attempt at a Solution



This is, as I've seen, not a new question. I'm good with how to solve it but where I see it should be tan(theta) = 4/3, I keep getting (as stated above) 4/3tan(theta) = 1 where if I were to bring the 4/3 to the other side I would end up with tan(theta) = 3/4.

My only problem is that I can't figure out how this 4/3 is ending up on the opposite side of the tan(theta). I don't know if I'm completely missing something or if it's just because it's 1:30 am...

Thanks
 
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aaaaaaaaaaaaaaaaaaaaaaand now I feel like a REAL genius...

tan is sin/cos... NOT cos/sin...

long time since trig + reading too quickly + no sleep = embarrassment by simple math (and negative signs)

sorry
 

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