Problems 23 through 28 explore the motion of projectiles under constant acceleration or deceleration.

What is the maximum height attained by the arrow of part (b) of Example 3?

(-a, 0) VR 13-axis US -US Example 3 (a,0) -x-axis UR FIGURE 1.2.5. A swimmer's problem (Example 4). Projectile motion (a) Suppose that a ball is thrown straight upward from the ground (yo 0) with initial velocity vo = 96 (ft/s, so we use g = 32 ft/s in fps units). Then it reaches its maximum height when its velocity (Eq. (16)) is zero, v (t) = -32t +96= 0, and thus when t = 3 s. Hence the maximum height that the ball attains is y (3) 32.32 +96.3+0= 144 (ft) (with the aid of Eq. (17)). (b) If an arrow is shot straight upward from the ground with initial velocity vo = 49 (m/s, so we use g = 9.8 m/s in mks units), then it returns to the ground when (9.8) + 49t = (4.9)t(t + 10) = 0, y(t) = and thus after 10 s in the air. A Swimmer's Problem Figure 1.2.5 shows a northward-flowing river of width w = 2a. The lines x = ta represent the banks of the river and the y-axis its center. Suppose that the velocity VR at which the water flows increases as one approaches the center of the river, and indeed is given in terms of distance x from the center by (-1/2). (18) You can use Eq. (18) to verify that the water does flow the fastest at the center, where UR = vo, and that UR = 0 at each riverbank. UR = Vo = Suppose that a swimmer starts at the point (-a, 0) on the west bank and swims due east (relative to the water) with constant speed us. As indicated in Fig. 1.2.5, his velocity vector (relative to the riverbed) has horizontal component vs and vertical component UR. Hence the swimmer's direction angle a is given by UR US Because tan = dy/dx, substitution using (18) gives the differential equation tan = dy 2- (-#) = dx US for the swimmer's trajectory y = y(x) as he crosses the river. (19)