The key idea is that, at launch, the object is given a horizontal component of velocity u_h, which does not change for the rest of the motion, and a vertical component of velocity u_v, which does change as the object accelerates in the (vertical) gravitational field.

A common technique in analysing projectiles is to deal with the vertical motion first, calculating the time of flight from one of

v_v = u_v - g t

s_v = u_vt - ½gt²

(ordinary SUVAT equations with up taken as positive) and then to calculate the horizontal range by substituting that time into

s_h = u_ht

The standard piece of bookwork

In the diagram on the right an object is launched with speed u at an angle q to the horizontal. The vertical component of its velocity is given by u_v = usinq, and by the time it reaches the ground again, it is back where it started from a vertical point of view, so that the vertical component of its displacement is zero. Substituting into the second of the equations above, we have           

0 = usinq t - ½gt²   

which easily yields, for time of flight 

t = 0 or (2 usinq  / g)

(The zero solution corresponds to the start, there being two moments when the vertical displacement is zero!)

Range = s_h = u_ht = ucosq (2 usinq  / g)

A little algebra and trigonometry produces

Range = (u²sin 2q )/ g

The maximum range occurs when sin2q has its maximum value of 1, which is when 2q = 90º. So the maximum range is achieved for a launch angle of 45º.

Other calculations

Substituting half of the time of flight into the second equation gives the maximum height to  which the projectile rises.  

If the projectile is launched horizontally, then the horizontal component of velocity is the launch velocity itself. Provided you are given the height of the cliff, or whatever, to use as s_v, you can still use the second equation to obtain the time of flight by putting u_v equal to zero.