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Examples in Chapter 3

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Problem 3.23 A man stands on the roof of a 150 m tall building and throws a rock with a velocity of 30 m/s at an angle of 330 above the horizontal. Ignore air resistance. Calculate: The maximum height above the roof reached by the rock The magnitude of the velocity of the rock just before it strikes the ground The horizontal distance from the base of the building to the point where the rock strikes the ground.

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Step 1: Draw It! 30 m/s Height, H 330 150 m Range, R

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**What do we know? The x- and y-components of the initial velocity**

Vx0=30*cos(330)=25.16 m/s Vy0=30*sin(330)=16.33 m/s The acceleration in the y-direction: ay=-g=-9.8 The acceleration in the x-direction: ax=0 The initial height of the rock, 150 m, = y0 The initial horizontal position of the rock, 0

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**Motion in the x- and y-directions**

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**At the top of the arc, vy(t)=0**

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**The magnitude of the velocity before it strikes ground.**

The rock strikes ground when y(t)=0

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The range of the rock The rock strikes ground after s

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Problem 3.31 In a test of a “g-suit” a volunteer is rotated in horizontal circle of radius 7.0 m. What is the period of rotation at which the centripetal acceleration has a magnitude of 3.0 g? 10.0 g?

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Step 1: Draw It! arad=3 g or 10 g 7.0 m

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What do we know? arad =3 g or 10 g R= 7.0 m Need to find T

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Plugging and Chugging arad =3 g or 10 g R= 7.0 m Need to find T

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Problem 3.37 A “moving sidewalk” in an airport moves at 1 m/s and is 35.0 m long. If a women steps on one end and walks at 1.5 m/s relative to the moving sidewalk, how much time does she require to reach the opposite side if She walks in the same direction as the sidewalk is moving? She walks against the motion of the sidewalk?

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Step 1: Draw It 1.5 m/s 1 m/s 1.5 m/s 1 m/s

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**Must find relative velocity**

Call the slide walk as reference frame A, therefore the woman’s velocity in this frame vA is 1.5 m/s Call a stationary observer frame of reference, B and the slidewalk’s velocity in this frame is vB =1.0 m/s The end points are fixed in reference B so I must adjust the woman’s velocity to reference B

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**Two different velocities**

If the woman and slide are in the same direction: Vp/B=Vp/A+VA/B Vp/A= velocity of woman relative to slidewalk=1.5 m/s VA/B= velocity of slidewalk relative to frame B=1.0 m/s Vp/B=1+1.5=2.5 m/s

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**Two different velocities cont’d**

If the woman and slide are in the opposite directions: Vp/B=Vp/A+VA/B Vp/A= velocity of woman relative to slidewalk=-1.5 m/s VA/B= velocity of slidewalk relative to frame B=1.0 m/s Vp/B=1-1.5=-0.5 m/s

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**Finally, In the same direction: vp/B=d/t where d=35 m and Vp/B=2.5 m/s**

t=35/2.5=14 s In the opposite direction: vp/B=d/t where d=35 m and Vp/B=0.5 m/s t=35/0.5=70 s

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Problem 3.58 A baseball thrown at an angle of 600 above the horizontal strikes a building 18 m away at a point 8 m above the point it is thrown. Ignore air resistance. Find the magnitude of the initial velocity of the baseball ( the velocity with which the baseball is thrown) Find the magnitude and direction of the velocity just before it strikes the building.

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Step 1: Draw It! v0 m/s 8 m 600 18 m

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**The Secret Weapon: The Trajectory Equation**

You can go through a lot of rigmarole but the most powerful tool in your projectile arsenal is this little formula below

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You know x=18 m y= 8 m a = 600 You need to find v0

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Part B) x=18 m, x0=0 y= 8 m, y0=0 a = 600 v0=16.55 m/s and v0x=16.55*cos(600)=8.275, voy=16.55*sin(600)=14.33 ax=0, ay=-9.8 m/s2 Need to find vx(t) and vy(t) when x=18 m

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3.3 Projectile Motion The motion of an object that is subjected to a constant, downward acceleration is termed projectile motion. We already covered 1D.

3.3 Projectile Motion The motion of an object that is subjected to a constant, downward acceleration is termed projectile motion. We already covered 1D.

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