First, find the formula for the time taken, t, for the projectile to travel the distance. Using the fact that the projectile reaches a velocity of zero at a time of 0.5t when at its maximum height and acceleration due to gravity is negative, the time of flight is dependent on vertical values so; v=u+at => 0=Vsinθ-g(0.5t) => t=(2Vsinθ)/g. Now for the range, also know as maximum displacement, substitute the time taken into the distance, x, formula with horizontal values; x=ut => x=Vcosθt => x(max)=Vcosθ((2Vsinθ)/g)) => x(max)=Range=(2sinθcosθ(V)^2)/g. Using trigonometric identity sin2θ=2sinθcosθ, we have Range=(sin2θ(V)^2)/g. To find the motion of the projectile, use the equation for displacement s=ut+0.5a(t)^2, therefore in vertical terms y=Vsinθ(t)-0.5g(t)^2, and thus insert the horizontal time taken which is derived from x=Vcosθt => t=x/Vcosθ, so the path the projectile follows on the x-y plane is y=Vsinθ(x/Vcosθ)-(0.5g(x)^2)/(Vcosθ)^2. Tidying this up, and using the fact that secθ=1/cosθ and tanθ=sinθ/cosθ; this means y=xtanθ-g((xsecθ)^2)/2(V)^2. Since the equation is in the form y=ax-bx^2, for some a,b, the motion of the projectile must be parabolic. And we are done.