- #1
emptymaximum
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nomenclature: vi = initial speed, z = height x,y are the 'ground', φ = elevation angle of the barrel, α. = α 'dot' = dα/dt, α: = double dot
show that an artillery cannon can hit a target anywhere in the surface given by:
g²r²=(vi²)²-2gz(vi²)
ignore air resistance. no wind.
note: this is not intended to be done as a surface integral.
so what i did was decided that it can shoot something directly above it:
zmax = vi²/2g, zD = 0
then considering the 2-d case (y=0)
and then as φ decreases, so rotates the point of max height (up to φ = ╥/2, but then the max height moves back inside this trajectory). This is going to make a paraboloid from (0,zmax) to (?,0). The equation of tha paraboloid is z = Ax + Bx² + C
at zmax x=0 so zmax = C = vi²/2g
z. = Ax. + Bx.²
z: = Ax: + Bx:² = -g, but x: = 0 so the equation for the paraboloid is wrong?
help please.
show that an artillery cannon can hit a target anywhere in the surface given by:
g²r²=(vi²)²-2gz(vi²)
ignore air resistance. no wind.
note: this is not intended to be done as a surface integral.
so what i did was decided that it can shoot something directly above it:
zmax = vi²/2g, zD = 0
then considering the 2-d case (y=0)
and then as φ decreases, so rotates the point of max height (up to φ = ╥/2, but then the max height moves back inside this trajectory). This is going to make a paraboloid from (0,zmax) to (?,0). The equation of tha paraboloid is z = Ax + Bx² + C
at zmax x=0 so zmax = C = vi²/2g
z. = Ax. + Bx.²
z: = Ax: + Bx:² = -g, but x: = 0 so the equation for the paraboloid is wrong?
help please.