Angle of Projection from Height and Range

AI Thread Summary
A projectile is launched such that its horizontal range is three times its maximum height, leading to the equation R = 3H. The relevant formulas for range and height are R = Vo²sin(2theta)/g and H = (Vo sin(theta)²)/(2g). The user initially struggles with the manipulation of these equations, specifically how to isolate tan(theta) and ends up with an incorrect expression. After some reflection, they realize their mistake stems from confusion between sine and cosine functions, compounded by fatigue. The discussion highlights the importance of careful mathematical manipulation and understanding of trigonometric identities.
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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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