Rich wrote in message
om...
Thanks for the replies.

However I have found out today that my orig equation T^2 ..... is
actually wrong.
Let me explain what I am after.

I am firing a projectile
I know the position it starts and its end position. The end position
will be either above or below the start position. This amount is
known. that is I know (Xinit, Yinit) amd (Xfinal, Yfinal)
I also know the angle that the projectile will be fired at.

What I would like to know are formulae that allow me to work out the
init firing velocity and the total time of flight.

I have looked at the links given in previous post and they all have
equations going the other way. That is you know the angle and the
firign velocity and they work out the final position and tiem of
flight.

Does anyone know these equations???

Thanks
Ricahrd

The physics are very simple: Two points, (t0, x0, y0) which is
the initial position and initial time, and, (t1, x1, y1) which is the final
position and final time. Let

X = x1 - x0
Y = y1 - y0
T = t1 - t0

The velocity in the x-direction is constant

Vx =Vx0

Therefore,

X = Vx0*T

The firing angle, A0, is given by

sin(A0) = Vy0 / V0
or Vy0 = V0*sin(A0)

cos(A0) = Vx0 / V0 = X / V0*T
or T = X / V0*cos(A0)

The acceleration of gravity, g, acts in the y-direction:

Y = Vy0*T - (g / 2)*T^2

Substituting for Vy0 and T from above yields

Y = X*tan(A0) - (g / 2)*T^2

Rearanging terms gives the desired equations for V0 and T:

T^2 = [ 2 / g ] [ X*tan(A0) - Y ]

V0 = X / T*cos(A0)

[Old Man]