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Figure 3.19 The nearly parabolic trajectories of a bouncing ball. We can get a lot of information from Eqs. (3.19) through (3.22). For example, the distance r from the origin to the projectile at any time t is Successive images of the ball are separated by equal time intervals. r = V + y? (3.23) Successive peaks decrease in height because the ball The projectile's speed (the magnitude of its velocity) at any time is loses energy with each bounce. v = Vu? + v (3.24) The direction of the velocity, in terms of the angle a it makes with the positive x-direction (see Fig. 3.17), is Uy tan a (3.25) The velocity vector v is tangent to the trajectory at each point. We can derive an equation for the trajectory's shape in terms of x and y by eliminating t. From Eqs. (3.19) and (3.20), we find t = x/(vocos ao) and y = (tan ao)x (3.26) 2u3 cos²c Figure 3.20 Air resistance has a large cumulative effect on the motion of a Don't worry about the details of this equation; the important point is its general form. Since vo, tan ao, cos ao, and g are constants, Eq. (3.26) has the form baseball. In this simulation we allow the baseball to fall below the height from which it was thrown (for example, the baseball could have been thrown from y = bx – cx a cliff). where b and c are constants. This is the equation of a parabola. In our simple model of projectile motion, the trajectory is always a parabola (Fig. 3.19). When air resistance isn't negligible and has to be included, calculating the trajec- tory becomes a lot more complicated; the effects of air resistance depend on velocity, so the acceleration is no longer constant. Figure 3.20 shows a computer simulation of the trajectory of a baseball both without air resistance and with air resistance pro- portional to the square of the baseball's speed. We see that air resistance has a very large effect; the projectile does not travel as far or as high, and the trajectory is no longer a parabola. y (m) Baseball's initial velocity: 100 E Vo = 50 m/s, ao = 53.1° 50 x (m) 300 100 200 -50 -100 With air No air resistance resistance

Figure 3.19 The nearly parabolic trajectories of a bouncing ball. We can get a lot of information from Eqs. (3.19) through (3.22). For example, the distance r from the origin to the projectile at any time t is Successive images of the ball are separated by equal time intervals. r = V + y? (3.23) Successive peaks decrease in height because the ball The projectile's speed (the magnitude of its velocity) at any time is loses energy with each bounce. v = Vu? + v (3.24) The direction of the velocity, in terms of the angle a it makes with the positive x-direction (see Fig. 3.17), is Uy tan a (3.25) The velocity vector v is tangent to the trajectory at each point. We can derive an equation for the trajectory's shape in terms of x and y by eliminating t. From Eqs. (3.19) and (3.20), we find t = x/(vocos ao) and y = (tan ao)x (3.26) 2u3 cos²c Figure 3.20 Air resistance has a large cumulative effect on the motion of a Don't worry about the details of this equation; the important point is its general form. Since vo, tan ao, cos ao, and g are constants, Eq. (3.26) has the form baseball. In this simulation we allow the baseball to fall below the height from which it was thrown (for example, the baseball could have been thrown from y = bx – cx a cliff). where b and c are constants. This is the equation of a parabola. In our simple model of projectile motion, the trajectory is always a parabola (Fig. 3.19). When air resistance isn't negligible and has to be included, calculating the trajec- tory becomes a lot more complicated; the effects of air resistance depend on velocity, so the acceleration is no longer constant. Figure 3.20 shows a computer simulation of the trajectory of a baseball both without air resistance and with air resistance pro- portional to the square of the baseball's speed. We see that air resistance has a very large effect; the projectile does not travel as far or as high, and the trajectory is no longer a parabola. y (m) Baseball's initial velocity: 100 E Vo = 50 m/s, ao = 53.1° 50 x (m) 300 100 200 -50 -100 With air No air resistance resistance

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please answer 3.19
Figure 3.19 The nearly parabolic trajectories of a bouncing ball. We can get a lot of information from Eqs. (3.19) through (3.22). For example, the distance r from the origin to the projectile at any time t is Successive images of the ball are separated by equal time intervals. r = V + y? (3.23) Successive peaks decrease in height because the ball The projectile's speed (the magnitude of its velocity) at any time is loses energy with each bounce. v = Vv? + vỷ (3.24) The direction of the velocity, in terms of the angle a it makes with the positive x-direction (see Fig. 3.17), is Uy tan a (3.25) The velocity vector v is tangent to the trajectory at each point. We can derive an equation for the trajectory's shape in terms of x and y by eliminating t. From Eqs. (3.19) and (3.20), we find t = x/(vocos ao) and y = (tan ao)x (3.26) 2u3 cos²c Figure 3.20 Air resistance has a large cumulative effect on the motion of a Don't worry about the details of this equation; the important point is its general form. Since vo, tan ao, cos ao, and g are constants, Eq. (3.26) has the form baseball. In this simulation we allow the baseball to fall below the height from which it was thrown (for example, the baseball could have been thrown from y = bx – cx a cliff). where b and c are constants. This is the equation of a parabola. In our simple model of projectile motion, the trajectory is always a parabola (Fig. 3.19). When air resistance isn't negligible and has to be included, calculating the trajec- tory becomes a lot more complicated; the effects of air resistance depend on velocity, so the acceleration is no longer constant. Figure 3.20 shows a computer simulation of the trajectory of a baseball both without air resistance and with air resistance pro- portional to the square of the baseball's speed. We see that air resistance has a very large effect; the projectile does not travel as far or as high, and the trajectory is no longer a parabola. y (m) Baseball's initial velocity: 100 E Vo = 50 m/s, ao = 53.1° 50 100 200 x (m) 300 -50 -100 With air No air resistance resistance
Transcribed Image Text:Figure 3.19 The nearly parabolic trajectories of a bouncing ball. We can get a lot of information from Eqs. (3.19) through (3.22). For example, the distance r from the origin to the projectile at any time t is Successive images of the ball are separated by equal time intervals. r = V + y? (3.23) Successive peaks decrease in height because the ball The projectile's speed (the magnitude of its velocity) at any time is loses energy with each bounce. v = Vv? + vỷ (3.24) The direction of the velocity, in terms of the angle a it makes with the positive x-direction (see Fig. 3.17), is Uy tan a (3.25) The velocity vector v is tangent to the trajectory at each point. We can derive an equation for the trajectory's shape in terms of x and y by eliminating t. From Eqs. (3.19) and (3.20), we find t = x/(vocos ao) and y = (tan ao)x (3.26) 2u3 cos²c Figure 3.20 Air resistance has a large cumulative effect on the motion of a Don't worry about the details of this equation; the important point is its general form. Since vo, tan ao, cos ao, and g are constants, Eq. (3.26) has the form baseball. In this simulation we allow the baseball to fall below the height from which it was thrown (for example, the baseball could have been thrown from y = bx – cx a cliff). where b and c are constants. This is the equation of a parabola. In our simple model of projectile motion, the trajectory is always a parabola (Fig. 3.19). When air resistance isn't negligible and has to be included, calculating the trajec- tory becomes a lot more complicated; the effects of air resistance depend on velocity, so the acceleration is no longer constant. Figure 3.20 shows a computer simulation of the trajectory of a baseball both without air resistance and with air resistance pro- portional to the square of the baseball's speed. We see that air resistance has a very large effect; the projectile does not travel as far or as high, and the trajectory is no longer a parabola. y (m) Baseball's initial velocity: 100 E Vo = 50 m/s, ao = 53.1° 50 100 200 x (m) 300 -50 -100 With air No air resistance resistance
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