A projectile is launched with initial speed v0 and angle θ over level ground. The projectile's maximum height H and horizontal range R are related by the equation R = 4H. Write an expression for the launch angle of the projectile.
The Attempt at a Solution
Knowing that maximum height H occurs at time .5t, so H = Vosin(θ) x .5t
Similarly horizontal range R occurs at time t, so R = Vocos(θ) x t
Multiplying both sides of H = Vyosin(θ) x .5t by 4 gives
4H = Vyosin(θ) x 2t
Because of the given R = 4H,
Vosin(θ) x 2t = Vocos(θ) x t.
All the Vo and t terms cancel leaving 2sin(θ) = cos(θ), or tan(θ) = 1/2.
Thus θ = arctan(1/2), but the answer given is arctan(1). I am off by a factor of 2 somewhere.
Thanks for your help!