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?
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...