Originally Posted by klm
would this be correct: y0=26 , v0=30, g=9.8 , t=4 ... so y=26+(30x4) .5(9.8)(4)^2 = 224.4 m ?
i got vy by doing v0sintheta= 26

No. In the formula I gave:
y0 is your initial position, which I presume is on the ground at height = 0
v0 (which I'll change to v0y) is the
vertical component of the initial velocity, what you call v0sin(theta)
g = 9.8 m/s^2
To make it less confusing, I'll relabel v0 to be v0y in my equation.