Ballistic problem: reachable region

[snellslaw]

Where did this equation come from? (attached)

I can only think of the equation for d found at the end of the section in this wikipedia article: http://en.wikipedia.org/wiki/Range_of_a_projectile#Flat_ground

Thanks!

 

[tiny-tim]

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

 

[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 = v_i²sin²(θ)/2g = v_i²/2g * (1 - cos²(θ))
Now if this were to match the equation in OP, then we need
v_i²cos²(θ)/2g = gR²/2v_i²
But subbing in R = v_i²sin(2θ)/g, we get
v_i²cos²(θ)/2g = v_i²sin²(2θ)/2g
But cos²(θ) =/= sin²(2θ)

 

[tiny-tim]

hi snellslaw!

Then y = v_i²sin²(θ)/2g = v_i²/2g * (1 - cos²(θ))

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

 

[snellslaw]

Thanks tiny-tim! :D
I think the line you quoted was not my question however;
we need v_i²cos²(θ)/2g = gR²/2v_i²
but this leads to cos²(θ) = sin²(2θ) which is not an equality.

Thanks again!