Ballistic problem: reachable region

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no, it looks like the 2D region of all points that the projectile can pass through with initial speed vo
 
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)
 
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hi snellslaw! :smile:
snellslaw said:
Then y = vi2sin2(θ)/2g = vi2/2g * (1 - cos2(θ))

where does this come from? surely y = 0 ? :confused:

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