- #1

gooby99

## Homework Statement

A projectile is launched with initial speed

*v*0 and angle θ over level ground. The projectile's maximum height

*H*and horizontal range

*R*are related by the equation

*R*= 4

*H*. Write an expression for the launch angle of the projectile.

## Homework Equations

R=4H

V

_{xo}=V

_{o}cos(θ)

V

_{yo}=V

_{o}sin(θ)

## The Attempt at a Solution

Knowing that maximum height H occurs at time .5t, so H = V

_{o}sin(θ) x .5t

Similarly horizontal range R occurs at time t, so R = V

_{o}cos(θ) x t

Multiplying both sides of H = V

_{yo}sin(θ) x .5t by 4 gives

4H = V

_{yo}sin(θ) x 2t

Because of the given R = 4H,

V

_{o}sin(θ) x 2t = V

_{o}cos(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!