A shot putter releases the shot some distance above the levelground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. [(c) How far did she throw the shot horizontally?] (d) Why does the expression for R in Example 3.8 not give the correct answer for part (c)? (e) How high was the shot above the ground when she released it? (f) Draw x-t, y-t, vx-t, and vy-t graphs for the motion EXAMPLE 3.8 Height and range of a projectile II: Maximum height, maximum rangeFigure 3.24 A launc"range. The range is shFind the maximum height h and horizontal range R (see Fig. 3.23) of aprojectile launched with speed u, at an initial angle ao between 0 and90°. For a given vo, what value of ao gives maximum height? Whatvalue gives maximum horizontal range?A4otheIDENTIFY and SET UP This is almost the same as parts (b) and (c) ofExample 3.7, except that now we want general expressions for h and R. Wealso want the values of ao that give the maximum values of h and R. In part(b) of Example 3.7 we found that the projectile reaches the high point ofits trajectory (so that v, = 0) at time t¡ = vo/g, and in part (c) we foundthat the projectile returns to its starting height (so that y = yo) at time2 = 20y/8 = 2t1. We'll use Eq. (3.20) to find the y-coordinate h at fị andEq. (3.19) to find the x-coordinate R at time 12. We'll express our answers interms of the launch speed vo and launch angle ao by using Eqs. (3.18).Launchangle:do = 3045°do = 60EXECUTE From Eqs. (3.18), vor = vocos ao and voy = vosinao. Hencewe can write the time 1 when v,2a0 = 90°, or aogiven initial speed ifO as%3DUbyUo sin aoEVALUATE Figure 3.trajectories of a ball p45°, and 60°. The initcases. The horizontalEquation (3.20) gives the height y = h at this time:%3DsinaoUosinaovf sin² aonearly the same for ugiven value of thean initial angle 90°resistance.)h = (tosinao)(indo) - ¿|(*28For a given launch speed u, the maximum value of h occurs forsina, = 1 and ao = 90°–that is, when the projectile is launchedstraight up. (If it is launched horizontally, as in Example 3.6, ao = 0and the maximum height is zero!)The time t2 when the projectile hits the ground isCAUTION Heightmemorizing the abo»only in the special ccan use the expressicheights are equal. TEthese equations do na20oy _ 200sinaoThe horizontal range R is the value of x at this time. From Eq. (3.19), this is20o sin aovổ sin 2aoR = (vocos ao)½ = (vo cos ao)KEYCONCEPT Wprojectile problems inumbers as far into t(We used the trigonometric identity 2 sin ao cos ao = sin 2ao, found inAppendix B.) The maximum valuesin 2a, is 1; this occurs whenexplore and understar

Name: A shot putter releases the shot some distance above the levelground wi
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Description: A shot putter releases the shot some distance above the levelground with a velocity of 12.0 m/s, 51.0° above the horizontal. The shot hits the ground 2.08 s later. Ignore air resistance. [(c) How far did she throw the shot horizontally?] (d) Why does