The general horizontal range formula for a projectile or how to win at shot-put

Neglecting air drag, the acceleration is g straight down, and thus the initial and final velocity vectors are related as shown

V_o

gt

V_f

Taking gt as the base of the triangle, its altitude is v_x , so its “area” in v² units is ½ v_xgt. But v_xt = x, the horizontal distance traveled. So x is twice the “area”/g. Using v_f as the base, the altitude is v_osin(q+f), where q is the angle of v_o above horizontal and f is the angle of v_f below horizontal. Putting this together, we get x = (v_ov_f /g)sin(q+f). You can easily see from this that the maximum x occurs when q+f = 90^o. Suppose we know v_o, q, and the initial and final y. How to find v_f and f: Take the starting point to be the origin. v_y² = v_yo² – 2gy , where y, the final vertical location is negative if below the start, as in the shot-put. Add v_x² to both sides of this equation and we have v_f² = v_o² – 2gy. (This could also be found using an energy approach.) To find f, v_x = v_ocosq = v_f cosf.

For maximum x, v_o and v_f are perpendicular, and you can show that tanq = v_o/v_f.