If a ball is launched at ground level at a velocity v and angle θ, find an expression for it's height at horizontal distance x.

Firstly it must be understood that vertical and horizontal motion can be treated as independent, as they are perpendicular to each other (and acceleration due to gravity only affects the y component of velocity). So considering vertical forces, an acceleration of -g (-9.81ms^-2) acts in the positive y direction. The vertical component of velocity can therefore be found with SUVAT (equations for constant acceleration). We know initial velocity in the positive y direction 'u_y' = vsin(θ), acceleration 'a' = -g, time is known as 't' and we are trying to find vertical displacement 'y' (using y instead of the normal 's' for displacement), so we can use y=u_yt + (1/2)at²=vsin(θ)t - (1/2)gt².
We know everything in this expression apart from t, so using that the horizontal velocity is constant at u_x=vcos(θ), use x=u_xt + (1/2)at²=vcos(θ)t + (1/2)at² , again where 'x' is horizontal displacement but where a=0 as there are no horizontal forces (so x=vcos(θ)t). Therefore we can rearrange for t=x/vcos(θ), then insert this into the y equation, so y=(vsin(θ)x)/vcos(θ) - (g/2)(x²/(vcos(θ))²), or more simply y=xtan(θ)-(g/2)(x/vcos(θ))².