Projectile Motion Max Height and Range

[gooby99]

Homework Statement

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.

Homework Equations

R=4H
V_xo=V_ocos(θ)
V_yo=V_osin(θ)

The Attempt at a Solution

Knowing that maximum height H occurs at time .5t, so H = V_osin(θ) x .5t
Similarly horizontal range R occurs at time t, so R = V_ocos(θ) x t
Multiplying both sides of H = V_yosin(θ) x .5t by 4 gives

4H = V_yosin(θ) x 2t

Because of the given R = 4H,

V_osin(θ) x 2t = V_ocos(θ) x t.

All the V_o 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!

 

[MarkFL]

You've neglected the fact that gravity is acting on the projectile.

I would begin with the acceleration resolved into the $x$ and $y$ components.

$a_y=-g$

$a_x=0$

Integrating with respect to time $t$, and using the initial velocity, what are your resolved velocities?