PHYSICS
PHYSICS
5th Edition
ISBN: 2818440038631
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 3, Problem 94P

(a)

To determine

The distance of the ball strikes the wall.

(a)

Expert Solution
Check Mark

Answer to Problem 94P

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

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=viyΔt12gΔt2 (IV)

Here, g is the gravitational acceleration.

Write the equation for change in height of the ball.

Δy=yfyi (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)Δt12gΔt2=(visinθ)Δxvicosθ12g(Δx2vi2cos2θ)=Δxtanθ(gΔx22vi2cos2θ)

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

yf=yi+Δy=yi+Δxtanθ(gΔx22vi2cos2θ)

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

yf=60cm(0.01m1cm)+(10m)tan10°((9.80m/s2)(10m)22(20m/s)2cos210°)=2.36m980m3/s2775.9m2/s2=2.36m1.26m=1.1m

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

(b)

To determine

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

(b)

Expert Solution
Check Mark

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=vfyviy (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).

vfyviy=gΔt

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

vfyvisinθ=g(Δxvicosθ)vfy=visinθ(gΔxvicosθ) (X)

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

vfy=(20m/s)sin10°((9.80m/s2)(10m)(20m/s)cos10°)=3.47m/s4.98m/s=1.51m/s1.5m/s

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

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