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how can I solve this for the angle A?
I have a projectile starting at the origin in two dimensional space with no gravity. It is fired at an angle A from the horizon with a constant speed S. It is fired at a target at position x, y that is moving with a speed p at an angle B from the horizon.
How can a derive a function for A in terms of B, p, s, x and y?
1 Answer
- Barry GLv 77 years agoFavorite Answer
The x and y components of the velocity of the projectile are u=ScosA and v=SsinA. If the projectile is launched from the origin at time t=0 then its co-ordinates at a later time t are (x,y) = (ut,vt).
The x and y components of the velocity of the target are w=PcosB and z=PsinB. If the projectile is launched at time t=0 from position (x0,y0) then its co-ordinates at a time t are (x,y) = (wt+x0,zt+y0).
The projectile will hit the target when the co-ordinates (x,y) of both are the same at the same time t, ie when
ut = wt+x0
vt = zt+y0.
We do not need to know t, so we can eliminate it :
(u-w)t = x0
(v-z)t = y0.
y0(u-w) = x0(v-z)
y0u-x0v = y0w-x0z
S(y0cosA-x0sinA) = P(y0cosB-x0sinB)
.
Comparing with the identity
Rcos(A+C) = RcosAcosC-RsinAsinC
we can write
RcosC=y0 and RsinC=x0 so that tanC=x0/y0. The last equations can be used to find C.
Therefore
Scos(A+C) = Pcos(B+C).
cos(A+C) = (P/S)cos(B+C).
The last equation can be used to find angle A.
