Is it possible to find the velocity of a projectile given a point ...

[hummingbird]

This one's been puzzling me for quite some time now. I've been given a point on the projectile's path as (140,6) and I know the angle is 55°, but that's all I have. I also know that at t=0, the height is 1m, if that helps.

[Jones]

that's not much info, but i guess that's the problem you're dealing with =]

[Jesse_J]

well, you just use the equation of trajectory and you will know the speed directly. the equation is:
y=(tanA*x)-[(g*x*x*)/(u*u*cosA*cosA*2)] where x and y are the co-ordinates and A is the angle of projection. you can easily derive this equation using simple equations of motion.

[ricki]

For the vertical direction we have:

y(t) = y(0) + v_y(0) t - 1/2 g t^2

For the horizontal direction:

x(t) = x(0) + v_x(0) t


This second equation implies:

t = ( x(t) - x(0) ) / v_x(0) .

Substituting that in the first gives the trajectory, y(x), which is the parabola :

y(x) = y(0) + v_y(0)/v_x(0)  ( x - x(0) ) - 1/2 g / (v_x(0)^2)  (x-x(0))^2

Now x(0) = 0 and y(0) = 1:

y = 1 + v_y(0)/v_x(0)  x  - g/(2 v_x(0)^2)  x^2 .

Differentiating wrt time gives:

v_y = v_y(0)   v_x / v_x(0) - g/( v_x(0))  x  


If we substitude (x,y) = (140,6) in these equations, we have two equations with four unknowns (initial and final velocity components). But we also know
 v_y/v_x = tan(55),
and from energy conservation
 1/2 m v^2 = 1/2 m v(0)^2 + m g (y-1)

So in principle it can be solved, as there are four equations with four unknowns.