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A shot putter releases the shot some distance above the level ground 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. (a) What are the components of the shot’s acceleration while in flight? (b) What are the components of the shot’s velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally?

A shot putter releases the shot some distance above the level ground 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. (a) What are the components of the shot’s acceleration while in flight? (b) What are the components of the shot’s velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally?

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A shot putter releases the shot some distance above the level ground 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. (a) What are the components of the shot’s acceleration while in flight? (b) What are the components of the shot’s velocity at the beginning and at the end of its trajectory? (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 lI: Maximum height, maximum range WITH VARIATION PROBLEMS Find the maximum height h and horizontal range R (see Fig. 3.23) of a projectile launched with speed y at an initial angle ao between 0 and 90°. For a given vọ, what value of ao gives maximum height? What value gives maximum horizontal range? Figure 3.24 A launch angle of 45° gives the maximum horizontal range. The range is shorter with launch angles of 30° and 60°. A 45° launch angle gives the greatest range; other angles fall shorter. IDENTIFY and SET UP This is almost the same as parts (b) and (c) of Example 3.7, except that now we want general expressions for h and R. We also 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 of its trajectory (so that v, = 0) at time t¡ = vos/8, and in part (c) we found that the projectile returns to its starting height (so that y = yo) at time 2 = 20oy/g = 2t. We'll use Eq. (3.20) to find the y-coordinate h at tj and Eq. (3.19) to find the x-coordinate R at time tɔ. We'll express our answers in terms of the launch speed vo and launch angle ao by using Eqs. (3.18). Launch angle: ao = 30° ao = 45° ao = 60° EXECUTE From Eqs. (3.18), vox = vo cos ao and voy = vosinɑo. Hence we can write the time t¡ when v, = 0 as 2a0 = 90°, or ao = 45°. This angle gives the maximum range for a given initial speed if air resistance can be ignored. Uoy Uo sin ao EVALUATE Figure 3.24 is based on a composite photograph of three trajectories of a ball projected from a small spring gun at angles of 30°, 45°, and 60°. The initial speed uo is approximately the same in all three cases. The horizontal range is greatest for the 45° angle. The ranges are nearly the same for the 30° and 60° angles: Can you prove that for a given value of v, the range is the same for both an initial angle ao and an initial angle 90° – ao? (This is not the case in Fig. 3.24 due to air resistance.) Equation (3.20) gives the height y = h at this time: ") - vf sin?ao ( sinao h = (vo sinao)( (sin ao 2g For a given launch speed up, the maximum value of h occurs for sina, = 1 and a, = 90°-that is, when the projectile is launched straight up. (If it is launched horizontally, as in Example 3.6, ¤o = 0 and the maximum height is zero!) The time t, when the projectile hits the ground is CAUTION Height and range of a projectile We don't recommend memorizing the above expressions for h and R. They are applicable only in the special circumstances we've described. In particular, you can use the expression for the range R only when launch and landing heights are equal. There are many end-of-chapter problems to which these equations do not apply. I 20oy 20o sinɑo The horizontal range R is the value of x at this time. From Eq. (3.19), this is 20o sin ao vổ sin 2ao R = (vocos ao)½ = (vo cos ao) KEYCONCEPT When you solve physics problems in general, and projectile problems in particular, it's best to use symbols rather than numbers as far into the solution as possible. This allows you to better explore and understand your result. (We used the trigonometric identity 2 sin ao cos ao = sin 2a0, found in Appendix B.) The maximum value of sin 2ao is 1; this occurs when
Transcribed Image Text:EXAMPLE 3.8 Height and range of a projectile lI: Maximum height, maximum range WITH VARIATION PROBLEMS Find the maximum height h and horizontal range R (see Fig. 3.23) of a projectile launched with speed y at an initial angle ao between 0 and 90°. For a given vọ, what value of ao gives maximum height? What value gives maximum horizontal range? Figure 3.24 A launch angle of 45° gives the maximum horizontal range. The range is shorter with launch angles of 30° and 60°. A 45° launch angle gives the greatest range; other angles fall shorter. IDENTIFY and SET UP This is almost the same as parts (b) and (c) of Example 3.7, except that now we want general expressions for h and R. We also 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 of its trajectory (so that v, = 0) at time t¡ = vos/8, and in part (c) we found that the projectile returns to its starting height (so that y = yo) at time 2 = 20oy/g = 2t. We'll use Eq. (3.20) to find the y-coordinate h at tj and Eq. (3.19) to find the x-coordinate R at time tɔ. We'll express our answers in terms of the launch speed vo and launch angle ao by using Eqs. (3.18). Launch angle: ao = 30° ao = 45° ao = 60° EXECUTE From Eqs. (3.18), vox = vo cos ao and voy = vosinɑo. Hence we can write the time t¡ when v, = 0 as 2a0 = 90°, or ao = 45°. This angle gives the maximum range for a given initial speed if air resistance can be ignored. Uoy Uo sin ao EVALUATE Figure 3.24 is based on a composite photograph of three trajectories of a ball projected from a small spring gun at angles of 30°, 45°, and 60°. The initial speed uo is approximately the same in all three cases. The horizontal range is greatest for the 45° angle. The ranges are nearly the same for the 30° and 60° angles: Can you prove that for a given value of v, the range is the same for both an initial angle ao and an initial angle 90° – ao? (This is not the case in Fig. 3.24 due to air resistance.) Equation (3.20) gives the height y = h at this time: ") - vf sin?ao ( sinao h = (vo sinao)( (sin ao 2g For a given launch speed up, the maximum value of h occurs for sina, = 1 and a, = 90°-that is, when the projectile is launched straight up. (If it is launched horizontally, as in Example 3.6, ¤o = 0 and the maximum height is zero!) The time t, when the projectile hits the ground is CAUTION Height and range of a projectile We don't recommend memorizing the above expressions for h and R. They are applicable only in the special circumstances we've described. In particular, you can use the expression for the range R only when launch and landing heights are equal. There are many end-of-chapter problems to which these equations do not apply. I 20oy 20o sinɑo The horizontal range R is the value of x at this time. From Eq. (3.19), this is 20o sin ao vổ sin 2ao R = (vocos ao)½ = (vo cos ao) KEYCONCEPT When you solve physics problems in general, and projectile problems in particular, it's best to use symbols rather than numbers as far into the solution as possible. This allows you to better explore and understand your result. (We used the trigonometric identity 2 sin ao cos ao = sin 2a0, found in Appendix B.) The maximum value of sin 2ao is 1; this occurs when
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