# Homework Help: Ballistic problem: reachable region

1. Jul 2, 2012

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2. Jul 2, 2012

### tiny-tim

no, it looks like the 2D region of all points that the projectile can pass through with initial speed vo

3. Jul 2, 2012

### snellslaw

Thanks! could you please explain further how you know this to be true?
If you look here: http://en.wikipedia.org/wiki/Trajectory#Range_and_height
I'm assuming x is the max range R?
Then y = vi2sin2(θ)/2g = vi2/2g * (1 - cos2(θ))
Now if this were to match the equation in OP, then we need
vi2cos2(θ)/2g = gR2/2vi2
But subbing in R = vi2sin(2θ)/g, we get
vi2cos2(θ)/2g = vi2sin2(2θ)/2g
But cos2(θ) =/= sin2(2θ)

Last edited: Jul 2, 2012
4. Jul 3, 2012

### tiny-tim

hi snellslaw!
where does this come from? surely y = 0 ?

if you put x = v2sin(2θ)/g into the equation, and θ = 45°, you do get y = 0

5. Jul 3, 2012

### snellslaw

Thanks tiny-tim! :D
I think the line you quoted was not my question however;
we need vi2cos2(θ)/2g = gR2/2vi2
but this leads to cos2(θ) = sin2(2θ) which is not an equality.

Thanks again!!

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