# Projectile Motion launch angles

1. Jan 18, 2014

### Jon.G

1. The problem statement, all variables and given/known data

A projectile is fired in such a way that its maximum height is a factor of 4 times its
horizontal range. Find the launch angle in degrees

2. Relevant equations
$h_{max}=\frac{V^{2}_{0}sin^{2}\theta}{2g} Range=\frac{V^{2}_{0}sin2\theta}{g}$

3. The attempt at a solution

I set h=4R giving
$sin^{2}(\theta)=8sin(2\theta);$
$sin^{2}(\theta)=16sin(\theta)cos(\theta);$
$sin(\theta)=16cos(\theta);$
$tan(\theta)=16$
$\theta=86^{o}$
but the answer is apparently 38.66 degrees.

So I assumed that I had misinterpreted the question, and in fact R=4h, however doing this I got an angle of 45 degrees.
I'm not sure where I've gone wrong

2. Jan 18, 2014

### Staff: Mentor

Hi Jon.G. Welcome to Physics Forums.

Given the problem statement as it stands your interpretation and calculations look fine to me (using a bit more precision, θ = 86.42°). Are you quoting the problem statement exactly? It could be that they changed the problem statement slightly, to make it a "new" question, but failed to change the answer key.

3. Jan 19, 2014

### Jon.G

I emailed my tutor and that is exactly what happened :)