Given a projectile is launched, from rest, at an angle θ and travels at a velocity V, what is the range and path of motion of the projectile? (Ignore air resistance.)

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.