Example Video Example ( A projectile is fired with an angle of elevation a and initial velocity Vo. (See the figure below.) 0 Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile. What value of a maximizes the range (the horizontal distance traveled)? Solution 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/s. Thus, a = -gj. Since v'(t) = a, we have v(t) = -gtj + C, where C = v(0) = v. Therefore, r'(t) = v(t) = -gtj + Vo Integrating again, we obtain r(t) = gt 2 ) + ( [ t But D = r(0) = 0, so the position vector of the projectile is given by r(t) = )+ + D. gt 2 ) + (+ t (1). If we write vol = 0 (the initial speed of the projectile), then vo= vo cos(a) i + sin() j and Equation (1) becomes r(t) (Vo cos(a))ti + [(V, sin(a)) - gi 0 The parametric equations of the trajectory are therefore x = (V cos(a))t y = (V sin(a))t - gt. The horizontal distance d is the value of x when y = 0. Setting y = 0, we obtain t = 0 or t = d = x = 2v sin(a) = g g sin(2x) g Clearly, d has its maximum value when sin(2x) = 1, that is, = 2 2 (20 sin(a)) This second value of t then gives g