1. Motion in Two or Three Dimensions
Chapter 3
Motion in Two or Three Dimensions

2. Vectors to represent the position of a body
Goals for Chapter 3
Vectors to represent the position of a body
Velocity vector using the path of a body
Acceleration vector of a body
Describe the curved path of projectile
Investigate circular motion
(Describe the velocity of a body as seen from different frames of reference)

3. Examples of projectile motion. Notice the effects of air resistance

4. Projectile Motion
A projectile is an object moving in two dimensions under the influence of Earth's gravity; its path is a parabola.

5. Projectile motion
Projectile is any body with initial velocity that follows path determined by gravity (& air resistance).
Begin by assuming “g” is constant (near ground), & neglecting air resistance, Earth’s curvature, rotation, & motion.

6. Projectile Motion
Understood by analyzing the horizontal and vertical motions separately.

7. Projectile Motion
Photo shows 2 balls starting to fall at the same time.
Yellow ball on right has an initial speed in the x-direction.
Red ball on left has vx0 = 0
Vertical positions of the balls are identical at identical times
Horizontal position of yellow ball increases linearly in time.

8. Projectile Motion
The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.

9. The x and y motion are separable—Figure 3.16
We can analyze projectile motion as horizontal motion with constant velocity and vertical motion with constant acceleration: ax = 0 and ay = g.

10. Solving Problems Involving Projectile Motion
Projectile motion is motion with constant acceleration in two dimensions,
Usually where vertical acceleration is g and coordinate system assumes UP = + motion!

11. Solving Problems Involving Projectile Motion
Read the problem carefully, and choose the object(s) you are going to analyze.
Draw a diagram.
Choose an origin and a coordinate system.
Decide on the time interval; this is the same in both directions, and includes only the time the object is moving with constant acceleration g.
Examine the x and y motions separately.

12. Solving Problems Involving Projectile Motion
6. List known and unknown quantities.
Remember that vx never changes, and that vy = 0 at the highest point.
7. Plan how you will proceed.
Use the appropriate equations; you may have to combine some of them, or find time from one set and use in the other

13. Solving Problems Involving Projectile Motion
Driving off a cliff
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff.
How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are?
Ignore air resistance.

14. Solving Problems Involving Projectile Motion
Example Driving off a cliff.
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
We WANT:
vx (m/s)

15. Solving Problems Involving Projectile Motion
Example Driving off a cliff.
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
We KNOW:
vy0 = 0 m/s!

16. Solving Problems Involving Projectile Motion
Example Driving off a cliff.
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
We KNOW:
Dy = -50 m!

17. Solving Problems Involving Projectile Motion
Example Driving off a cliff.
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
We KNOW:
Dx = +90 m!

18. Solving Problems Involving Projectile Motion
Example Driving off a cliff.
A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the cliff top to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
We KNOW:
ay = -9.8!
ax = 0!

19. A body projected horizontally
A motorcycle leaves a cliff horizontally.

20. Solving Problems Involving Projectile Motion
For Horizontal:
Dx = +90 m
Dx = vxt + ½ axt2
but ax = 0 so
Dx = vxt
vx = Dx /t
So vx = 90/t
How do we get t ??

21. Solving Problems Involving Projectile Motion
Get t from Vertical:
Dy = -50 m
Dy = vyt + ½ ayt2
but ax = and v0y = 0 so
Dy = -50 = - ½ gt2
So t = [-2(-50)/9.8] ½ = 3.2 sec

22. Solving Problems Involving Projectile Motion
Get v0x from time:
Dx = +90 m
vx = 90/t
= 90/3.2 = 28.2 m/sec
=~ 62 mile/hour

23. Projectile Motion
If object is launched at initial angle θ0 with the horizontal, analysis is similar except that the initial velocity has a vertical component.

24. Solving Problems Involving Projectile Motion
Example: A kicked football.
A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.
It travels through the air (assume no resistance) and lands at the ground.

25. Example: A kicked football.
Calculate
the maximum height,
the time of travel before the football hits the ground,
how far away it hits the ground.
the velocity vector at the maximum height,
the acceleration vector at maximum height.

26. Solving Problems Involving Projectile Motion
STEP 1: Coordinates! (and signs!)
+x horizontally to the right
+y vertically upwards
acceleration of gravity = -9.8 m/s/s

27. Solving Problems Involving Projectile Motion
STEP 2: COMPONENTS! (and signs!)
+vx0 horizontally to the right = v0 cos q
+vy0 vertically upwards = v0 sinq
ay acceleration of gravity = -9.8 m/s/s only in y

28. Solving Problems Involving Projectile Motion
STEP 3: LOOK FOR 3 THINGS!
vf = vi + at
Dx = ½ (vi + vf)*t
Dx = vi*t + ½ at2
Dx = vf*t – ½ at2
Vf2 = Vi2 +2aDx

29. Solving Problems Involving Projectile Motion
In x direction, ax = 0!
vfx = vix
Dx = vix*t = v cosq * t

30. Solving Problems Involving Projectile Motion
In y direction, ay = -9.8!
vfy = viy – 9.8t = v0 sin q - gt
Dy = ½ (viy + vfy)*t
Dy = viy*t – 4.9t2 = (v0 sin q) t - ½ gt2
Dx = vfy*t + 4.9t2
vfy2 = viy2 – 19.6Dy = (v0 sin q)2 - 2gDy

31. Solving Problems Involving Projectile Motion

32. Solving Problems Involving Projectile Motion
Solve it!
vx = 16 m/s
vy = 12 m/s
Max height = 7.3 m
Max distance in x (“range”) = 39 m
Time = 2.5 seconds

33. Example: Level horizontal range.
The horizontal range is defined as the horizontal distance the projectile travels before returning to its original height (which is typically the ground); that is, y(final) = y0.
(a) Derive a formula for the horizontal range R of a projectile in terms of its initial speed v0 and angle θ0.

34. Solving Problems Involving Projectile Motion
Example: Level horizontal range
(b) Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m/s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?

35. Solving Problems Involving Projectile Motion
Example: A kicked football.
A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.
It travels through the air (assume no resistance) and lands at the ground.

36. Solving Problems Involving Projectile Motion
A football is kicked at an angle θ0 = 37.0° with a velocity of 20.0 m/s, as shown.
Suppose the football was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground?

37. Solving Problems Involving Projectile Motion
Suppose the football was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground?
Dy = - 1 meter = vyt + ½ ayt2

38. A good problem to try algebraically…
A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height h above the ground.
What is the maximum launch speed you could give this projectile if you shot it straight up? Express your answer in terms of h and g .
Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part A. At what maximum angle above the horizontal should you launch the projectile?

39. Solving Problems Involving Projectile Motion
The wrong strategy!
A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.

40. Tranquilizing a falling monkey
Where should the zookeeper aim?

41. Solving Problems Involving Projectile Motion
Projectile motion is parabolic:
Taking the equations for x and y as a function of time, and combining them to eliminate t, we find y as a function of x:
This is the equation for a parabola.

42. The equations for projectile motion
If we set x0 = y0 = 0, the equations describing projectile motion are shown at the right.
The trajectory is a parabola.

43. Height and range of a projectile
A baseball is batted at an angle.
Vo = 37.0 m/s at seconds where is it?

44. Maximum height and range of a projectile
What initial angle will give the maximum height and the maximum range of a projectile?

45. The effects of air resistance
Calculations become more complicated.
Acceleration is not constant.
Effects can be very large.
Maximum height and range decrease.
Trajectory is no longer a parabola.

46. Position vector
The position vector from the origin to point P has components x, y, and z.

47. Average velocity
Average velocity between two points = displacement divided by time
Same direction as the displacement.

48. Instantaneous velocity
Instantaneous velocity is instantaneous rate of change of position vector with respect to time.
Components of instantaneous velocity are vx = dx/dt, vy = dy/dt, and vz = dz/dt.
Instantaneous velocity of a particle is always tangent to its path.

49. Average acceleration
The average acceleration during a time interval t is defined as the velocity change during t divided by t.

50. Average acceleration
The average acceleration during a time interval t is defined as the velocity change during t divided by t.

51. Instantaneous acceleration
Instantaneous acceleration is the instantaneous rate of change of the velocity with respect to time.
Any particle following a curved path is accelerating, even if it has constant speed.
Components are ax = dvx/dt, ay = dvy/dt, and az = dvz/dt.

52. Direction of the acceleration vector
Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

53. Direction of the acceleration vector
Direction of acceleration vector depends on whether speed is constant, increasing, or decreasing,

54. Uniform circular motion
For uniform circular motion, the speed is constant and the acceleration is perpendicular to the velocity.

55. Acceleration for uniform circular motion
For uniform circular motion, the instantaneous acceleration always points toward the center of the circle and is called the centripetal acceleration.
The magnitude of the acceleration is arad = v2/R.
The period T is the time for one revolution, and arad = 4π2R/T2.

56. Nonuniform circular motion
If the speed varies, the motion is nonuniform circular motion.
The radial acceleration component is still arad = v2/R, but there is also a tangential acceleration component atan that is parallel to the instantaneous velocity.

57. Acceleration of a skier
Skier moving on a ski-jump ramp.
Figure shows the direction of the skier’s acceleration at various points.

58. Relative velocity
The velocity of a moving body seen by a particular observer is called the velocity relative to that observer, or simply the relative velocity.
A frame of reference is a coordinate system plus a time scale.

59. Relative velocity in one dimension
If point P is moving relative to reference frame A, we denote the velocity of P relative to frame A as vP/A.
If P is moving relative to frame B and frame B is moving relative to frame A, then the x-velocity of P relative to frame A is vP/A-x = vP/B-x + vB/A-x.

60. Relative velocity on a straight road
Motion along a straight road is a case of one-dimensional motion.

61. Relative velocity in two or three dimensions
We extend relative velocity to two or three dimensions by using vector addition to combine velocities.
A passenger’s motion is viewed in the frame of the train and the cyclist.

62. Flying in a crosswind
A crosswind affects the motion of an airplane.