The distance of the ball strikes the wall.

Question

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Answer 1

Question

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

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Answer 2

Question

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

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Answer 3

Question

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

Answer 6

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

Answer 7

Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

Answer 8

(a)

Expert Solution

Answer to Problem 94P

The distance of the ball strikes the wall is $1.1 m$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the equation of motion for the ball taken to reach the wall.

$Δx=vxΔt$ (I)

Here, $Δx$ is the displacement of the ball, $vx$ is the velocity of the ball along x axis, and $Δt$ is the time taken by the ball to reach the wall.

Write the equation of motion for the ball’s initial velocity along x axis.

$vx=vicosθ$ (II)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball’s initial velocity along y axis.

$viy=visinθ$ (III)

Here, $vi$ is the initial velocity of the ball and $θ$ is the vertical angle above the ground.

Write the equation of motion for the ball reaches certain height.

$Δy=viyΔt+12ayΔt2$

Here, $Δy$ is the change in height of the ball, $viy$ is the initial velocity of the ball along y axis, and $ay$ is the acceleration of the ball.

Rewrite the above equation for gravitational acceleration.

$Δy=viyΔt−12gΔt2$ (IV)

Here, $g$ is the gravitational acceleration.

Write the equation for change in height of the ball.

$Δy=yf−yi$ (V)

Here, $yf$ is the final height and $yi$ is the initial height.

Conclusion:

Substitute equation (II) in equation (I) to find $Δt$ .

$Δx=(vicosθ)ΔtΔt=Δxvicosθ$ (VI)

Substitute equation (III) and (VI) in equation (IV) to find $Δy$ .

$Δy=(visinθ)Δt−12gΔt2=(visinθ)Δxvicosθ−12g(Δx2vi2cos2θ)=Δxtanθ−(gΔx22vi2cos2θ)$

Substitute the above relation in equation (V) to find $yf$ .

$yf=yi+Δy=yi+Δxtanθ−(gΔx22vi2cos2θ)$

Substitute $60 cm$ for $yi$ , $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in the above equation.

$yf=60 cm(0.01 m1 cm)+(10 m)tan10°−((9.80 m/s2)(10 m)22(20 m/s)2cos210°)=2.36 m−980 m3/s2775.9 m2/s2=2.36 m−1.26 m=1.1 m$

Therefore, the distance of the ball strikes the wall is $1.1 m$ .

Answer 13

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

Answer 14

(b)

To determine

Whether the ball rolls up or down when it hits the wall.

Answer 15

(b)

Expert Solution

Answer to Problem 94P

The ball rolls down when it hits the wall.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the equation for change in velocity of the ball along y axis.

$Δv=vfy−viy$ (VII)

Here, $vfy$ is the final velocity and $viy$ is the initial velocity.

Write the equation of motion for the acceleration of ball.

$Δv=ayΔt$ (VIII)

Here, $Δv$ is the change in velocity.

Rewrite the equation (VIII) for gravitational acceleration.

$Δv=−gΔt$ (IX)

Conclusion:

Compare equation (VII) and (IX).

$vfy−viy=−gΔt$

Replace $visinθ$ for $viy$ and $Δxvicosθ$ for $Δt$ in above relation.

$vfy−visinθ=−g(Δxvicosθ)vfy=visinθ−(gΔxvicosθ)$ (X)

Substitute $10 m$ for $Δx$ , $9.80 m/s2$ for $g$ , $20 m/s$ for $vi$ , and $90°−80°=10°$ for $θ$ in equation (X).

$vfy=(20 m/s)sin10°−((9.80 m/s2)(10 m)(20 m/s)cos10°)=3.47 m/s−4.98 m/s=−1.51 m/s≃−1.5 m/s$

Since, $vfy<0$ , so that ball is on its way down.

Answer 20

Ch. 3.1 - Conceptual Practice Problem 3.1 Bank Balance When...Ch. 3.1 - Prob. 3.2PP Ch. 3.2 - Two displacements and have x- and y-components...Ch. 3.2 - Prob. 3.3PP Ch. 3.2 - Prob. 3.2BCP Ch. 3.3 - Prob. 3.4PP Ch. 3.4 - Prob. 3.4ACP Ch. 3.4 - Prob. 3.4BCP Ch. 3.4 - Prob. 3.5PP Ch. 3.4 - Prob. 3.4CCP

The distance of the ball strikes the wall.

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The distance of the ball strikes the wall.

Answer to Problem 94P

Explanation of Solution

Whether the ball rolls up or down when it hits the wall.

Answer to Problem 94P

Explanation of Solution

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