the range of a projectile
The range of a projectile is how far it travels horizontally before coming back down to its launch height — the distance from where a thrown ball leaves your hand to where it lands, measured along the ground. It is the practical question an archer, a long-jumper, or an artillery gunner most wants answered: given how hard and at what angle I launch, how far will it go?
Precisely, for a projectile launched over flat ground at speed v_0 and angle theta above the horizontal (ignoring air resistance), the range is R = v_0^2 sin(2 theta) / g. Because the horizontal velocity is constant, R is just the horizontal speed multiplied by the total time aloft. Two things follow: range grows with the square of the launch speed (double the speed, quadruple the range), and it depends on angle through sin(2 theta), which is largest when 2 theta = 90 degrees — that is, at a launch angle of 45 degrees.
A neat consequence is that pairs of angles that add to 90 degrees give the same range: 30 and 60 degrees land the same distance away (one via a low fast arc, the other via a high slow one), because sin(2 theta) is the same for both. This clean formula is an idealization, though: it assumes no air resistance and level ground, launch and landing at the same height. With air drag the true maximum-range angle is somewhat less than 45 degrees, and launching from a cliff or into a valley changes the answer entirely.
Launch a ball at v_0 = 20 m/s and theta = 45 degrees on flat ground. Range R = (20^2) sin(90 degrees) / 9.8 = 400 x 1 / 9.8 = about 41 m. At 30 degrees the same speed gives R = 400 x sin(60 degrees) / 9.8 = about 35 m — and 60 degrees gives that same 35 m.
Maximum range on flat ground is at 45 degrees; complementary angles share a range.
The 45-degree-for-maximum-range and R = v_0^2 sin(2 theta) / g results assume no air resistance and equal launch and landing heights. Throw from a raised platform, or add real air drag, and the best angle shifts below 45 degrees.