A small projectile is fired vertically downward into a fluid medium with an initial velocity of 60 m/s. Due to the drag resistance of the fluid the projectile experiences a deceleration of a = (-0.4) m/s, where vis in m/s. Determine the projectiles velocity and position 4s after it is fired. SOLUTION Coordinate System. Since the motion is downward, the position coordinate is positive downward, with origin located at 0, Fig. 12-3. Velocity. Here a w) and so we must determine the velocity as a function of time using a = du/dt, since this equation relates v, a, and (Why not use v = w, ta?) Separating the variables and integrating with no 60 m/s when I = 0, yields (+) a = -0.4 Fig. 12-3 0 m/ 0.4 -04(42) ? . -1-0 0.8l 8- (60) . -for 1-1/2 + 0.8 ms Here the positive root is taken, since the projectile will continue to move downward. When I = 4s v = 0.559 m/s! Position. Knowing v - ), we can obtain the projectiles position from v - ds/dtsince this equation relates so, and t. Using the initial conditions - 0, when I = 0, we have (+) -- ia - Sorosi 0.8L(60 any - or on ds + 0.8 dt (60)? + + 0.4 ( m When I = 4s 5 - 4.43m