Eliminate t from the two equations
x = vtcos(theta)
and
y = vtsin(theta) - (gt^2/2)
and obtain a relationship between x and y (assume that v, g and theta are constants)
from the first:
t = x/(vcosØ)
from the 2nd:
y = vsinØ(x/(vcosØ) - (g/2)(x^2/(v^2 cos^2 Ø)
y = xtanØ - (gx^2)/(2v^2cos^2 Ø)
You can also read the excellent article on "trajectory" in wikipedia, where these equations are derived.
To eliminate t from the two equations x = vtcos(theta) and y = vtsin(theta) - (gt^2/2), we need to isolate t in one of the equations and then substitute it into the other equation. Let's start by isolating t in the first equation:
x = vtcos(theta)
Divide both sides of the equation by vcos(theta):
x / (vcos(theta)) = t
Now we have an expression for t in terms of x, v, and theta. We can substitute this expression in the second equation:
y = vtsin(theta) - (gt^2/2)
Substituting t with x / (vcos(theta)):
y = v(x / (vcos(theta)))sin(theta) - (g(x / (vcos(theta)))^2/2)
Simplifying this expression:
y = xsin(theta) - (gx^2sin^2(theta)) / (2v^2cos^2(theta))
Now we have a relationship between x and y:
y = xsin(theta) - (gx^2sin^2(theta)) / (2v^2cos^2(theta))
This is the relationship between x and y obtained by eliminating t from the two equations.