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given:

- let theta be: B

-equation for:

(E1)horizontal component is x = vxt; where vx= v0cosB

(E2)vertical component is y= y0 +vyt - (1/2)gt(squared); where vy = v0sinB

using these components an equation of initial velocity{v0} = R{range of projectile}/ square root of (2y/g) is derived

the problem is i can't figure out the steps on how to come up with this particular equation of initial velocity using the given components

knowing that according to galileo "an object projected horizontally will reach the ground in the same time as an object is dropped vertically", i came up with these equations

derived from (E1): t= x/vx (E3)

i then substituted E3 in E2:

y = yo + vy (x/vx) - (1/2)g(x/vx)^2

then, when i plug in the values ( vy = v0sinB, vx= v0cosB), i ended up with this equation:

y = yo + v0sinB (x/v0cosB) - (1/2)g(x/v0cosB)^2

i assumed that yo=o, because it was not found in the final equation (initial velocity eq.):

y= (xv0sinB/v0cosB)- ((gx^2)/2v0^2cos^2B))

2v0^2cos^2B(y= (xv0sinB/v0cosB)- ((gx^2)/2v0^2cos^2))}2v0^2cos^2B

y2v0^2cos^2B= (x2v0cosBv0sinB) - gx^2

i don't know where to go from here...

please help, it will be greatly appreciated