Standard form: y = xtan(theta) - gx^$\frac{2}{2u^2*cos^2(theta}$)

Alternative form using Range: y = x*tan(theta)(1 - x/R) where R = $u^2$*sin$\frac{2*theta}{g}$

This form is elegant because:

• At x = 0: y = 0 (launch point) ✓
• At x = R: y = 0 (landing point) ✓
• At x = R/2: y = $\frac{R}{4}$*tan(theta) = H (maximum height) ✓

Radius of curvature at highest point: At the top, v = ucos(theta), centripetal acceleration = g $r_{top}$ = $v^2$/g = $u^2$cos^2$$\frac{theta}{g}

Radius of curvature at launch: $r_{launch}$ = u^$\frac{2}{g*cos(theta}$) [velocity is u, component of g perpendicular to v is g*cos(theta)]