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Match the answers with questions based on Example 3.10 (I attached pictures of example 3.10 from the textbook)  1.) ball's initial speed 2.) ball's initial speed along the x-component         3.) ball's initial speed along the y-component 4.) ball's initial velocity 5.) ball's acceleration along the y-component 6.) ball's acceleration along the x-component   A. 23 m/s, 22° from the horizontal B. 9.8 m/s2 C. -9.8 m/s2 D. 8.6 m/s E. 23 m/s F. 0 G. 21.3 m/s

Match the answers with questions based on Example 3.10 (I attached pictures of example 3.10 from the textbook)  1.) ball's initial speed 2.) ball's initial speed along the x-component         3.) ball's initial speed along the y-component 4.) ball's initial velocity 5.) ball's acceleration along the y-component 6.) ball's acceleration along the x-component   A. 23 m/s, 22° from the horizontal B. 9.8 m/s2 C. -9.8 m/s2 D. 8.6 m/s E. 23 m/s F. 0 G. 21.3 m/s

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Match the answers with questions based on Example 3.10 (I attached pictures of example 3.10 from the textbook) 

1.) ball's initial speed

2.) ball's initial speed along the x-component        

3.) ball's initial speed along the y-component

4.) ball's initial velocity

5.) ball's acceleration along the y-component

6.) ball's acceleration along the x-component

 

A.

23 m/s, 22° from the horizontal
B.

9.8 m/s2
C.

-9.8 m/s2
D.

8.6 m/s
E.

23 m/s
F.

0
G.

21.3 m/s
3.10 A free kick In a soccer free kick, a player kicks a stationary ball toward the goal that is 18 m away. He kicks the ball at an angle of 22° from the horizontal at a speed of 23 m/s. How long does the ball take to reach the goal? And how far off the ground is the ball when it reaches the goal? STRATEGIZE This is a projectile motion problem with the initial velocity at an angle above the horizontal. This means that we will have both vertical and horizontal components of the initial velocity when we apply the equations of Problem-Solving Approach 3.1 O. PREPARE Following Problem-Solving Approach 3.1 O, we prepare the visual overview shown in Figure 3.31 O. In choosing our axes, we've placed the origin at the point where the ball is kicked. The initial velocity vector is tilted at 22° above the horizontal, so the components of the initial velocity are (v.)i= v; cos 0 = (23 m/s)(cos 22°) = 21.3 m/s (v,);= v; sin 0 = (23 m/s)(sin 22°) = 8.6 m/s FIGURE 3.31 Visual overview of a kicked soccer ball. Known Xi = 9i = 0m, ti= Os Vi = 23 m/s, 0 = 22° Xf = 18 m Xf Find X;o Yio ti Xf> Yfo tf
Transcribed Image Text:3.10 A free kick In a soccer free kick, a player kicks a stationary ball toward the goal that is 18 m away. He kicks the ball at an angle of 22° from the horizontal at a speed of 23 m/s. How long does the ball take to reach the goal? And how far off the ground is the ball when it reaches the goal? STRATEGIZE This is a projectile motion problem with the initial velocity at an angle above the horizontal. This means that we will have both vertical and horizontal components of the initial velocity when we apply the equations of Problem-Solving Approach 3.1 O. PREPARE Following Problem-Solving Approach 3.1 O, we prepare the visual overview shown in Figure 3.31 O. In choosing our axes, we've placed the origin at the point where the ball is kicked. The initial velocity vector is tilted at 22° above the horizontal, so the components of the initial velocity are (v.)i= v; cos 0 = (23 m/s)(cos 22°) = 21.3 m/s (v,);= v; sin 0 = (23 m/s)(sin 22°) = 8.6 m/s FIGURE 3.31 Visual overview of a kicked soccer ball. Known Xi = 9i = 0m, ti= Os Vi = 23 m/s, 0 = 22° Xf = 18 m Xf Find X;o Yio ti Xf> Yfo tf
Xf Pina Xio Yis ti Xf, Yfs tf Yos tf SOLVE The problem-solving approach suggests using one component of the motion to solve for At. We can do so using the kinematic equation for the horizontal position, Xf= ¤i + (v,)iAt We know that x¡ = 0 m, so we can solve this equation for At to get 18 m At = 0.845 s 21.3 m/s where we have kept one extra significant figure for our calculations. We can now use the kinematic equation for the vertical position to find the height of the ball when it reaches the goal: Yi + (v,);At –¿g(At)² Om + (8.6 m/s) (0.845 s) – (9.8 m/s³) (0.845 s)² = Yf = 3.8 m The ball is 3.8 m off the ground when it reaches the goal. ASSESS A height of 3.8 m-about 12 ft-seems reasonable for the height of a kicked ball. Unfortunately, the height of the goal itself is only 2.44 m, so this kick sails easily over the crossbar.
Transcribed Image Text:Xf Pina Xio Yis ti Xf, Yfs tf Yos tf SOLVE The problem-solving approach suggests using one component of the motion to solve for At. We can do so using the kinematic equation for the horizontal position, Xf= ¤i + (v,)iAt We know that x¡ = 0 m, so we can solve this equation for At to get 18 m At = 0.845 s 21.3 m/s where we have kept one extra significant figure for our calculations. We can now use the kinematic equation for the vertical position to find the height of the ball when it reaches the goal: Yi + (v,);At –¿g(At)² Om + (8.6 m/s) (0.845 s) – (9.8 m/s³) (0.845 s)² = Yf = 3.8 m The ball is 3.8 m off the ground when it reaches the goal. ASSESS A height of 3.8 m-about 12 ft-seems reasonable for the height of a kicked ball. Unfortunately, the height of the goal itself is only 2.44 m, so this kick sails easily over the crossbar.
Expert Solution
Step 1

The velocity has a magnitude and a direction, the magnitude is called speed.

The horizontal and vertical components of the velocity are shown in figure below.

Advanced Physics homework question answer, step 1, image 1

In projectile motion the x and y directions are independent, as gravity acts only vertically the x direction is unaffected, hence in horizontal direction the particle moves with constant velocity whereas vertically it experiences acceleration due to gravitation.

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