You will never need to use that formula. Having said that, it can be a nice shortcut if projectiles aren't your strong point. One negative thing with it is that if it is projected from a different height to where it lands, that formula doesn't work. So I would suggest instead knowing how to derive it.
All you have to do is find the time of flight (by making y=0 or whatever the final height of the projectile is, and then solving for t), and then substitute that into R=uxt.
In the case of the height of the projectile being the same at the beginning and end of its flight, you do this by:
y = uyt -0.5*g*t2
0 = uyt -0.5*g*t2
0 = t(uy-0.5*g*t)
Therefore t=0, or t=2*uy/g
since t=0 is the time of projection, the time of flight is given by t=2uy/g
Substituting this into R=uxt gives:
R=ux(2uy/g)
R=2uxuy/g
Since ux=Vcos@, and uy=Vsin@,
R=2(Vcos@)(Vsin@)/g
R=V2*2cos@sin@/g
Since sin(2@) = 2sin@cos@,
R=V2sin(2@)/g, which is your equation.
In the equation:
V is the initial velocity of the projectile (not purely horizontal or vertical, but a combination of both)
@ is the angle of projection, e.g. if it is fired at 15o, you would enter @ as 15o, and sin(2@) = sin(30o)
g is the acceleration due to gravity, which on in this course is given as 9.8m.s-1 on Earth.
