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Examples in Chapter 3

Examples in Chapter 3. 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 33 0 above the horizontal. Ignore air resistance. Calculate: The maximum height above the roof reached by the rock

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Examples in Chapter 3

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  1. Examples in Chapter 3

  2. 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.

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

  4. 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

  5. Motion in the x- and y-directions

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

  7. The magnitude of the velocity before it strikes ground. • The rock strikes ground when y(t)=0

  8. The range of the rock • The rock strikes ground after 4.111 s

  9. 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?

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

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

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

  13. 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?

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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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.

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

  21. 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

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

  23. 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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