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

2. Projectile Motion
Velocity of an object can be broken into two independent components, vertical and horizontal (normally the horizontal is “x” and vertical is “y” in the Cartesian coordinate system)
Since the two components are independent they do not affect each other
Vector quantities, e.g. velocity and acceleration, have magnitude and direction. Scalar quantities (mass and speed) have only magnitude

3. Projectile Motion
It can be understood by analyzing the horizontal and vertical motions separately.

4. Question
What hits the ground first an object moving with a large horizontal speed or one that is just dropped from the same height?
The demonstration showed that they hit at the same time, i.e. motion in the x and y directions are independent!

5. Projectile Motion
The speed in the x-direction is constant; in the y-direction the object moves with constant acceleration g.
This photograph shows two balls that start to fall at the same time. The one on the right has an initial speed in the x-direction. It can be seen that vertical positions of the two balls are identical at identical times, while the horizontal position of the yellow ball increases linearly.
The Physics Classroom

6. A dropped object is a projectile

7. Projectiles
An object thrown in the air is a projectile

8. Projectiles
Many projectiles not only undergo a vertical motion, but also undergo a horizontal motion. That is, as they move upward or downward they are also moving horizontally. There are the two components of the projectile's motion - horizontal and vertical motion.

9. Question
What will be the acceleration of a rock thrown straight upward at the moment it reaches the tippity-top of its trajectory?
a = 9.8 m/s2

10. Question
A 1-kg rock is thrown at 10 m/s straight upward. Neglecting air resistance, what is the net force that acts on it when it is half way to the top of its path?
9.8 Newtons

11. Horizontally Launched Projectile
Motion
Vertical
Forces
(Present? - Yes or No) (If present, what dir'n?) No
Yes The force of gravity acts downward
Acceleration
Yes "g" is downward at 9.8 m/s/s
Velocity
(Constant or Changing?)
Constant
Changing (by 9.8 m/s each second)

12. Non-Horizontally Launched Projectiles
Once more, the presence of gravity does not affect the horizontal motion of the projectile. The projectile still moves the same horizontal distance in each second of travel as it did when the gravity switch was turned off. The force of gravity is a vertical force and does not affect horizontal motion; perpendicular components of motion are independent of each other.

13. Questions
Consider a plane moving with a constant speed at an elevated height above the Earth's surface. In the course of its flight, the plane drops a package from its luggage compartment. What will be the path of the package and where will it be with respect to the plane? And how can the motion of the package be described?

14. As can be seen from the animation, the package follows a parabolic path and remains directly below the plane at all times. As the package falls, it undergoes a vertical acceleration; that is, there is a change in its vertical velocity. This vertical acceleration is attributed to the downward force of gravity which acts upon the package.

15. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane?
below the plane and behind it.
directly below the plane
c. below the plane and ahead of it

16. Example
A rescue plane needs to drop supplies to stranded hikers on a ridge 235m below. The plane is traveling with a horizontal velocity of 69.4 m/s (250km/hr).
a) How far in advance of the hikers must the supplies be dropped?
b) If the plane need to drop the supplies 425m from the hikers what vertical velocity does the package need to be given.
c) In part b what will be the velocity of the package when it lands

17. Answers First determine the time t = √(2d/g) = √(470m / 9.8m/s2)
= 6.9 seconds
Next determine distance
d = velocity * time
= 69.4 m/s * 6.9 seconds
= 480 meters

18. Answers b) First determine the time t = distance / velocity
t = 425m / 69.4m/s
= 6.1 seconds
Determine Vy
-235m = Vy*t - ½ 9.8 t2
Vy = -235/t + 4.9*t
= -38.5m/s m/s
= -8.6 m/s

19. Answers c) Determine the two velocity components Vx = 69.4 m/s
Vy = -8.6 m/s - 9.8m/s2 *6.1s
= m/s
Use the Pythagorean theorem
V = √(Vx2 + Vy2)
V = 97.4 m/s

20. Flight time
Once the time to rise to the peak of the trajectory is known, the total time of flight can be determined. For a projectile which lands at the same height which it started, the total time of flight is twice the time to rise to the peak.

21. Hang Time What is the hang time of good leapers in the NBA?
The best leapers have vertical jumps of ~ 1.25meter
d = ½ gt2
t = 2X√(2d/g)
= 0.9 seconds

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

23. Question True or False:
The velocity at the top of the trajectory is zero.
False, horizontal velocity does not change, the vertical velocity is zero.

24. Projectile Velocity Components
Projectile Velocity Components
This is to say that the vertical velocity changes by 9.8 m/s each second and the horizontal velocity never changes. This is indeed consistent with the fact that there is a vertical force acting upon a projectile but no horizontal force. A vertical force causes a vertical acceleration - in this case, an acceleration of 9.8 m/s/s.
Time
Horizontal
Velocity
Vertical
0 s
20 m/s, right
1 s
9.8 m/s, down
2 s
19.6 m/s, down
3 s
29.4 m/s, down
4 s
39.2 m/s, down
5 s
49.0 m/s, down

25. The diagram depicts an object launched upward with a velocity of 75
The diagram depicts an object launched upward with a velocity of 75.7 m/s at an angle of 15 degrees above the horizontal. For such an initial velocity, the object would initially be moving 19.6 m/s, upward and 73.1 m/s, rightward.
Time
Horizontal
Velocity
Vertical
0 s
73.1 m/s, right
19.6 m/s, up
1 s
9.8 m/s, up
2 s
0 m/s
3 s
9.8 m/s, down
4 s
19.6 m/s, down
5 s
29.4 m/s, down
6 s
39.2 m/s, down
7 s
49.0 m/s, down

26. A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

27. One more look
y = 0.5 • g • t2
x = vix • t

28. Question
The boy on the tower throws a ball 20 meters downrange as shown. What is his pitching speed?
20 m/sec

29. Kinematics
The two sets of three equations blow are the kinematic equations which will be used to solve projectile motion problems.

30. Range of projectiles
The range (horizontal distance traveled) of a projectile depends on the initial velocity and the launch angle. Ignoring air resistance, the maximum distance is obtained for a launch angle of 45 degrees.
Note, for a given velocity of launch, the range is the same for angles that add to 90 degrees.

31. Example
A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
Horizontal Info:
Vertical Info:
x = 35.0 m
y = m
vix = ???
viy = 0 m/s
ax = 0 m/s/s
ay = -9.8 m/s/s
Use y = viy • t • ay • t2 to solve for time; the time of flight is 2.12 seconds.
Now use x = vix • t • ax • t2 to solve for vix, note that ax is 0 m/s/s.
By substituting 35.0 m for x and 2.12 s for t, the vix can be found to be 16.5 m/s.
                                                                                          

32. Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal.

33. Example
A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

34. Answers t = 2*Viy/g 3.61 s = t Horizontal Component Vertical Component
vix = vi•cos(Theta) vix = 25 m/s•cos(45)
vix = 17.7 m/s
viy = vi•sin(Theta) viy = 25 m/s•sin(45 deg)
viy = 17.7 m/s
Horizontal Information
Vertical Information
x = ??? vix = 17.7 m/s
vfx = 17.7 m/s
ax = 0 m/s/s
y = ??? viy = 17.7 m/s
vfy = m/s
ay = -9.8 m/s/s
t = 2*Viy/g
3.61 s = t

35. Answers x = (17.7 m/s)•(3.6077 s) Thus,x = 63.8 m
The horizontal displacement of the projectile is 63.8 m.
y = (17.7 m/s)•(1.80 s) + 0.5*(-10 m/s/s)•(1.80 s)2
Using a calculator, this equation can be simplified to
y = 31.9 m + (-15.9 m)
And thus, y = 15.9 m

37. Order
When added together in this different order, these same three vectors still produce a resultant with the same magnitude and direction as before (22 m, 310 degrees). The order in which vectors are added using the head-to-tail method is insignificant.

38. Vector Addition
All vectors can be broken down into x and y components…then we can add the x and y components individually and find the resultant using the Pythagorean theorem.

39. Finding direction
Once the measure of the angle is determined, the direction of the vector can be found. In this case the vector makes an angle of 45 degrees with due East. Thus, the direction of this vector is written as 45 degrees. ( the direction of a vector is the counterclockwise angle of rotation which the vector makes with due East.)

40. Pythagorean practice R2 = (5)2 + (10)2 R2 = 125 R = SQRT (125)
R = 11.2 km
R2 = (30)2 + (40)2
R2 = 2500
R = SQRT (2500)
R = 50 km

41. Direction Convention
The direction of a vector is often expressed as an angle of rotation of the vector about its "tail" from either east, west, north, or south. For example, a vector can be said to have a direction of 40 degrees North of East.
Note 60 degrees south of west is 240 degrees from East

42. Barb Dwyer recently submitted her vector addition homework assignment
Barb Dwyer recently submitted her vector addition homework assignment. As seen below, Barb added two vectors and drew the resultant. However, Barb Dwyer failed to label the resultant on the diagram. For each case, which is the resultant (A, B, or C)? Explain.
Diagram A: A is the resultant of B + C. Barb added B + C using the head-to-tail method and then drew the resultant from the tail of the first vector (B) to the head of the last vector (C).
Diagram B: A is the resultant of C + B. Barb added B + C using the head-to-tail method a and then drew the resultant from the tail of the first vector (C) to the head of the last vector (B).
                                                                                                            

43. The above diagram shows what is occasionally a difficult concept to believe. Many students find it difficult to see how 10 N + 10 N could ever be equal to 10 N

44. Resolution of forces

45. On two different occasions during a high school soccer game, the ball was kicked simultaneously by players on opposing teams. In which case (Case 1 or Case 2) does the ball undergo the greatest acceleration? Explain your answer
Case 2 results in the greatest acceleration. Even though the individual forces are greater in Case 1, the net force is greatest in Case 2. Acceleration depends on the net force; it is not dependent on the size of the individual forces.

46. Problem
The answer is 866 N, upward.
The vertical component can be found if a triangle is constructed with the 1000 N diagonal force as the hypotenuse. The vertical component is the length of the side opposite the hypotenuse. Thus,
sin (60 degrees) = (Fvert) / (1000 N)
Solving for Fvert will give the answer 866 N.
Consider the tow truck at the right. If the tensional force in the cable is 1000 N and if the cable makes a 60-degree angle with the horizontal, then what is the vertical component of force which lifts the car off the ground?

47. Equilibrium
If an object is at equilibrium, then the forces are balanced. Balanced is the key word which is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s.

48. If an object is at rest and is in a state of equilibrium, then we would say that the object is at "static equilibrium."
Force A B C
Magnitude
3.4 N
9.2 N
9.8 N
Direction
161 deg.
70 deg.
270 deg

49. Example
Suppose the tension in both of the cables is measured to be 50 N and that the angle which each cable makes with the horizontal is known to be 30 degrees. What is the weight of the sign?
Weight = 2*25 N = 50N

50. There is an important principle which emanates from some of the trigonometric calculations performed above. The principle is that as the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases. To illustrate this, consider a 10-Newton picture held by three different wire orientations as shown in the diagrams below. In each case, two wires are used to support the picture; each wire must support one-half of the sign's weight (5 N). The angle which the wires make with the horizontal is varied from 60 degrees to 15 degrees. Use this information and the diagram below to determine the tension in the wire for each orientation.
19.3 N
7.1 N
5.8 N.

51. The sign below hangs outside the physics classroom, advertising the most important truth to be found inside. The sign is supported by a diagonal cable and a rigid horizontal bar. If the sign has a mass of 50 kg, then determine the tension in the diagonal cable which supports its weight.
The tension is 980 Newtons.

52. Consider the situation below in which a force is directed at an angle to the horizontal. In such a situation, the applied force can be resolved into two components. These two components can be considered to replace the applied force at an angle. By doing so, the situation simplifies into a familiar situation in which all the forces are directed horizontally and vertically.

53. Question
To test your understanding, analyze the two situations below to determine the net force and the acceleration.
The net force is 69.9 N, right and the acceleration is 3.5 m/s/s, right.
Note that the vertical forces balance but the horizontal forces do not. The net force
Fnet = N, right - 60 N, left = 69.9 N, right
The mass ism = (Fgrav / g) = 20 kg
So the acceleration is
a = (69.9 N) / (20 kg) =3.50 m/s/s.

54. Inclined Planes
An object placed on a tilted surface will often slide down the surface. The rate at which the object slides down the surface is dependent upon how tilted the surface is; the greater the tilt of the surface, the faster the rate at which the object will slide down it. In physics, a tilted surface is called an inclined plane. Objects are known to accelerate down inclined planes because of an unbalanced force. To understand this type of motion, it is important to analyze the forces acting upon an object on an inclined plane.

55. Inclined planes
The first peculiarity of inclined plane problems is that the normal force is not directed vertically.
The force of gravity will be resolved into two components of force - one directed parallel to the inclined surface and the other directed perpendicular to the inclined surface.

56. Inclined planes
The perpendicular component of the force of gravity is directed opposite the normal force and as such balances the normal force. The parallel component of the force of gravity is not balanced by any other force.
In the absence of friction and other forces the acceleration of an object on an incline is the value of the parallel component (m*g*sinθ) divided by the mass (m).

57. Example
With no friction fill in the blanks for the two Diagrams of roller coaster cars
6930
4900
9800
9800
6.93
8.49
8490
6930

58. Friction and Inclined Planes
In the presence of friction or other forces the situation is slightly more complicated. Consider the diagram shown at the right. The perpendicular component of force still balances the normal force since objects do not accelerate perpendicular to the incline. Yet the frictional force must also be considered when determining the net force.

59. This problem (and all inclined plane problems) can be simplified through a useful trick known as "tilting the head." An inclined plane problem is in every way like any other net force problem with the sole exception that the surface has been tilted. Thus, you transform the problem back into the form with which you are more comfortable. Once the force of gravity has been resolved into its two components and the inclined plane has been tilted, the problem should look very familiar. Merely ignore the force of gravity (since it has been replaced by its two components) and solve for the net force and acceleration.

60. Example
As an example consider the situation depicted in the diagram at the right. The free-body diagram shows the forces acting upon a 100-kg crate which is sliding down an inclined plane. The plane is inclined at an angle of 30 degrees. The coefficient of friction between the crate and the incline is 0.3. Determine the net force and acceleration of the crate.

61. Example
Begin the above problem by finding the force of gravity acting upon the crate and the components of this force parallel and perpendicular to the incline. The force of gravity is 980 N and the components of this force are:
849
490N
849N
980
Fparallel = 490 N (980 N • sin30 )
Fperpendicular = 849 N (980 N • cos30 ).

62. Example Step 2
Now the normal force can be determined to be 849 N (it must balance the perpendicular component of the weight vector).
The force of friction can be determined from the value of the normal force and the coefficient of friction;
849
255
980
(Ffrict = μFnorm= 0.3 • 849 N).
Ffrict is 255 N

63. Answers
The net force is the vector sum of all the forces. The forces directed perpendicular to the incline balance; the forces directed parallel to the incline do not balance. The net force is
Fnet = 235 N (490 N N).
The acceleration is 2.35 m/s/s
(Fnet/m = 235 N/100 kg).
849
255
490N
849N
980

64. Hint a = 0, since speed is constant.
In the following diagram, a 100-kg box is sliding down a frictional surface at a constant speed of 0.2 m/s. The incline angle is different in each situation. Analyze each diagram and fill in the blanks.
Hint a = 0, since speed is constant.
849
Fgrav= 100kg*9.8m/s2 = 980N
490
F┴ = Fgrav*cos 30 = 849N
490N
F┴ = Fnorm = 849N
849N
F║ = Fgrav*sin30= 490N
Ffrict= F║ = 490N
980

65. Vector Addition in one dimension
Current 6 miles/hour
Boat 10 miles/hour
Resultant: 16 miles/hour
A boat goes downstream at 10 miles/hour
with a current of 6 miles/hour. The result is
That the boat moves at 16 miles/hour
relative to the riverbank.

66. Vector addition in one dimension
Boat 10 miles/hour
Current 6 miles/hour
Resultant: 4 miles/hour
A boat goes upstream at 10 miles/hour against a
current of 6 miles/hour. The result is that the boat
moves at 4 miles/hour relative to the riverbank.

67. Example
For the boat traveling at 10 miles/hour relative to the river, and the current of the river at 6 miles/hour.
If the boat’s destination downstream is 8 miles from it’ starting point how long does it take to get there?
How long does the return trip take?

68. Note, we have solved part of the problem
We have found the velocity of the boat relative to the riverbank in our vector addition discussion
A boat goes downstream at 10 miles/hour with a current of 6 miles/hour. The result is that the boat moves at 16 miles/hour relative to the riverbank
A boat goes upstream at 10 miles/hour against a current of 6 miles/hour. The result is that the boat moves at 4 miles/hour relative to the riverbank

69. Answers Moving downstream Time = distance/velocity
= 8 miles / (16 miles/hour)
= ½ hour
Moving Upstream
= 8 miles / (4 miles/hour)
= 2 hours

70. Plane
On occasion objects move within a medium which is moving with respect to an observer. For example, an airplane usually encounters a wind - air which is moving with respect to an observer on the ground below.

72. Suppose that the river was moving with a velocity of 3 m/s, North and the motor boat was moving with a velocity of 4 m/s, East. What would be the resultant velocity of the motor boat .
(4.0 m/s)2 + (3.0 m/s)2 = R2
16 m2/s2 + 9m2/s2 = R2
25 m2/s2 = R2
SQRT (25 m2/s2) = R
5.0 m/s = R
Boat
tan (theta) = (opposite/adjacent) tan (theta) = (3/4)
theta = invtan (3/4)
theta = 36.9 degrees

73. Example
A motor boat traveling 4 m/s, East encounters a current traveling 7.0 m/s, North.
What is the resultant velocity of the motor boat?
If the width of the river is 80 meters wide, then how much time does it take the boat to travel shore to shore?
What distance downstream does the boat reach the opposite shore?

74. Answers
The resultant velocity can be found using the Pythagorean theorem. The resultant is the hypotenuse of a right triangle with sides of 4 m/s and 7 m/s. It is
Magnitude = SQRT [ (4 m/s)2 + (7 m/s)2 ] = 8.06 m/s
Direction = invtan [ (7 m/s) / (4 m/s) ] = 60°
The time to cross the river is
t = d / v = (80 m) / (4 m/s) = 20 s
3. The distance traveled downstream is
d = v • t = (7 m/s) • (20 s) = 140 m

75. TRUE or FALSE:
A boat heads straight across a river. The river flows north at a speed of 3 m/s. If the river current was greater, then the time required for the boat to reach the opposite shore would not change.

76. Vector Addition: Forces
A pack of five Arctic wolves are exerting five different forces upon the carcass of a 500-kg dead polar bear. A top view showing the magnitude and direction of each of the five individual forces is shown in the diagram at the right.

77. Vector Addition The net force is 39.4 N at 324 degrees.
The acceleration of the polar bear is m/s/s. This value is found by dividing Fnet by m (39.4 N) / (500 kg)