We set up the axes so that the projectile starts at the origin. since the force due to gravity acts downward, we have f = ma = −mgj where g = |a| ≈ 9.8 m/s2. thus a = −gj. since v'(t) = a, we have v(t) = incorrect: your answer is incorrect. + c where c = v(0) = v0. therefore r'(t) = v(t) = −gtj + v0. integrating again, we obtain r(t) = + v0 + d. but d = r(0) = 0, so the position vector of the projectile is given by r(t) = + v0. (1) if we write |v0| = v0 (the initial speed of the projectile), then v0 = i + v0 sin(α) j and equation (1) becomes r(t) = (v0 cos(α))ti + [(v0 sin(α))t − 1 2 gt2]j. the parametric equations of the trajectory are therefore x = (v0 cos(α))t y = (v0 sin(α))t − 1 2 gt2. the horizontal distance d is the value of x when y = 0. setting y = 0, we obtain t = 0 or t = (2v0 sin(α))/g. this second value of t then gives d = x = 2v0 sin(α) g = v02 g = v02 sin(2α) g . clearly, d has its maximum value when sin(2α) = 1, that is, α =