1. Projectile HW #1: Handout
Projectile Motion
Projectile HW #1: Handout

2. Projectile Motion:
Introduction: Projectile motion refers to freefall motion in ______ dimensions. The motion of the object will have a ______________ component and a _______________ component.
There is a constant acceleration due to gravity, , points downwards only. Here the g refers to:
two
horizontal
vertical
downwards
downwards
There are two coordinates to describe projectile motion. The _____ coordinate and component refer to _____________ motion and the _____ coordinate and component refer to _____________ motion. x
horizontal y
vertical
We need to define the same concepts used in Ch. 2 for our study of two dimensional motion.

3. Position: An object must be given a location in space
Position: An object must be given a location in space. A two dimensional coordinate system is used:
+y = up
The object’s location can be described in relation to the origin. The origin can be chosen to be any place convenient.
The position can be represented by a vector whose coordinates are (x,y).
+x = right, or any other horizontal direction
The coordinates (x,y) also represent the x and y components of the vector position.

4. Displacement: The displacement of an object is the ____________ of ____________ of an object. The displacement is a ____________! The displacement is represented as:
change
position
vector +y
Initial position +x
Final position

5. Average velocity: The average velocity of an object represents the __________ at which position ____________.
rate
changes
The average velocity is the displacement of the object divided by the ______________ time.
elapsed
To simplify the equations, we will always take the initial time to be __________ seconds.
zero
With this change, the ______ time always equals the ________ time.
final
elapsed

6. Instantaneous velocity: The instantaneous velocity of an object represents the __________ at one ____________ of time. The instantaneous velocity is represented by the letter v, and it is also a vector.
velocity
instant Vx θ
To find the resultant, apply Pythagoras and tan-1 θ Vy

7. Average acceleration: The average acceleration of an object represents the __________ at which velocity ____________.
rate
changes
velocity
The average acceleration is the change of the ___________ of the object divided by the ______________ time.
elapsed
Instantaneous acceleration: The instantaneous acceleration of an object represents the _____________ at one ____________ of time. The instantaneous acceleration is represented by the letter a, and it is also a vector.
acceleration
instant

8. Uniformly Accelerated Motion: This kind of motion has a constant ________________. Since the acceleration is constant, the _____________ and the _________________ acceleration are equal.
acceleration
average
instantaneous
Freefall: The acceleration of an object will be due to gravity only. Gravity pulls with a constant acceleration towards the ground, or _____________ downward. The acceleration can be written as a vector as follows:
vertically
As before, g just represents the numeric value of the acceleration of gravity.
The – sign shows the direction points downwards!

9. 2 – dimensional motion difficulty: The motion of a projectile in two dimensions is quite complex!
The way to solve these problems is to resolve all the motion into components.

10. All the motion completely separates into components
All the motion completely separates into components. The motion in the x direction is independent of the motion in the y direction.
By separating the motion into x and y components, the motions become independent of one another. The motion in the x – direction is not affected by the motion in the y – direction.

11. Equations of Motion: The motion of a projectile in each dimension can be represented by the uniform motion equations from Ch. 2. Apply the equations to each coordinate axis separately. The general equations for uniform motion from Ch. 2 are:

12. Apply these equations to each coordinate axis
Apply these equations to each coordinate axis. For the x – direction, the acceleration is _________. The equations from Ch. 2 can be adapted to 2 –dimensional motion by adding subscripts to the variables that are vectors. The subscripts give the component name:
zero
The acceleration in the x – direction, ax, is zero.
Dx is the horizontal displacement.
vox is the initial velocity in the x direction.
vx is the final velocity.
The x component of the velocity is constant and distance is rate times time.

13. ay = – g
For the y – direction, the acceleration is _________. Substitute y for x and get the equations of motion for the vertical part of the motion:

14. The motion in the x – direction and the y – direction are independent of one another. The motion of a projectile is that of a constant velocity in the x – direction and a uniformly accelerated motion in the y – direction.

15. To solve projectile problems, just follow the basic guidelines given in Ch. 2. Read the problem, draw a picture, write down what is given and unknown, choose equations, and solve.
Example #1: A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above a flat, horizontal beach, as shown below. a. How long after being released does the stone strike the beach below the cliff?
b. How far from the base of the cliff does the stone land? c. With what speed and angle of impact does the stone land?

16. vox
a. Use the y – direction to find the time to fall.
Dy = – 50.0 m voy = 0

17. solve for t:
b. Use the x – direction to find the horizontal distance.

18. c. Solve for each component of the final velocity just before impact.
For the x – direction:
For the y – direction:
Note: The minus sign on vy is to show it points downwards!

19. vx = m/s
Use Pythagorean theorem: q
vy = – 31.3 m/s v
Use inverse tangent:
below horizontal

20. Example #2: A ball is launched horizontally from a height h above the ground. At the same moment, an identical ball is dropped from rest from the same height. Which ball will hit the ground first?
Solution: Both balls hit the ground at the same time. For the ball launched horizontally, the motion in the x – direction and the motion in the y – direction separate. The motion in the y – direction only depends on the y – direction, and is independent of the x – direction. That means the ball dropped straight down and the ball launched horizontally have the same vertical motion. They should arrive at the bottom at the same time with the same vertical speed.
Video demonstration:

21. Example #3: A hunter aims his banana cannon directly at a monkey hanging on a branch. If the monkey lets go of the branch at the moment the cannon is fired, will the monkey catch the banana?
Both banana and monkey are accelerated equally gravity.
No matter how slow the banana is fired, or how far the monkey falls, the two will always make contact.
This can be shown a with a couple of different scenarios. First what would happen if gravity were “turned off”?

22. If gravity were “off”, the banana would travel directly to the monkey.

23. If gravity is restored, and the cannon points directly to the monkey, the banana will still arrive to the monkey. Both fall the same distance due to gravity as they move. Here is a high projectile speed example:

24. Here is a low projectile speed example:

25. Here is a high projectile speed aimed too high:

26. Example #4: A student stands at the edge of a building and throws a stone horizontally over the edge with a speed of 12.0 m/s. The stone lands 2.41 seconds after it is thrown. a. How tall is the building?
b. How far from the base of the building does the ball land? c. With what speed and angle of impact does the stone land?
Look for an equation with Dy and t and solve.
vox = 12.0 m/s
voy = 0 m/s
yo = h = ?
y = 0
Dy = 0 – h

27. b. Solve for the x – direction displacement:
c. Solve for each component of the final velocity just before impact.
For the x – direction:
For the y – direction:
The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

28. vx = m/s
Use Pythagorean theorem: q
vy = – 23.6 m/s v
Use inverse tangent:
below horizontal

29. Example #5: Bubba stands at the edge of a building and throws an opossum horizontally over the edge with a speed of 6.50 m/s. The opossum lands 24.1 meters horizontally from the base of the building. How tall is the building? How long for the opossum to land? With what speed and angle of impact does the opossum land?
You do not have to solve the questions in the order given. Start with the x – direction and solve for time.
vox = 6.50 m/s
voy = 0 m/s
yo = h = ?
y = 0
Dy = 0 – h
Dx = 24.1 m

30. Now that the time is given, follow the last example to find height and final velocity.
Look for an equation with Dy and t and solve.

31. Solve for each component of the final velocity just before impact.
For the x – direction:
For the y – direction:
The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

32. vx = m/s
Use Pythagorean theorem: q
vy = – 36.3 m/s v
Use inverse tangent:
below horizontal

33. Example #6: Bubba stands at the edge of a building and throws an armadillo horizontally over the edge with a speed of 10.0 m/s. The armadillo lands on the ground with a final velocity directed at 55.0o below the horizontal. What is the height of the building?
Think of what is given to solve…..
vox = 10.0 m/s
voy = 0 m/s
yo = h = ?
y = 0
Dy = 0 – h
vy = ?

34. Next use the vy to solve for the height:

35. Projectile Motion Day 2 & 3
Projectile HW #2

36. Projectiles launched at an angle to the horizontal: Yesterday’s notes involved projectiles launched horizontally. In this case the projectiles had a horizontal component to the velocity but not a vertical component. Today the initial velocity will have a magnitude, vo, and an initial angle, qo. The components to the initial velocity are as follows:
Angle above the horizontal…..
The x – component is adjacent, so use cosine. v vy q vx so
The y – component is opposite, so use sine. so

37. Angle below the horizontal…..vx
The x – component is adjacent, so use cosine. q vy v so
The y – component is opposite, so use sine. This component also points downwards, so it is negative. so

38. Example #7: A ball is launched at 25. 0 m/s at an initial angle of 36
Example #7: A ball is launched at 25.0 m/s at an initial angle of 36.9o above the horizontal. (a) What are the x and y components of the initial velocity? vo
voy qo
vox

39. (b) What is the greatest height reached by the ball?+
voy = m/s
vy,top = 0 m/s
Dy = ? -
Recall: Vo = 25.0 m/s

40. Note: This greatest height can be written as:
At what angle should the object be thrown to reach the greatest height?
Reason: The largest value that sin(q) can have is +1, and that occurs only when the angle is 90o.

41. (c) How long did it take the ball to reach the highest point?
voy = m/s
vy,top = 0 m/s
t = ?

42. (d) What was the total time the ball was in the air?
voy = m/s
yo = 0 m
yf = 0 m
Dy = 0 m
t = ?
Think of the ball being thrown up vertically with Voy only.

43. Note: The total time in the air is just twice the time to the top
Note: The total time in the air is just twice the time to the top. This is the same result as in chapter 2. There is the same symmetry here as in the purely vertical motion in Ch. 2: When an object starts and stops at the same vertical height, the time to travel to the top is equal to the time to fall back down.

44. (e) What is the final velocity of the ball?
Start with the x – component:
For the y – direction:
The last step is to combine these into magnitude and direction. Draw the components and use Pythagorean theorem and inverse tangent.

45. vx = + vocosqo
Use Pythagorean theorem: q
vy = – vosinqo v
Use inverse tangent:
below horizontal

46. (f) What is the range of the ball?
The range of a projectile is the horizontal distance the particle travels.
(g) Write a general formula for the range using only vo, qo, and g.

47. Start substituting values in:

48. Note that 45o gives the greatest range.
From your Trig class, you learned (or will learn) of an identity:
Note that 45o gives the greatest range.

50. Example #8: A football is kicked upwards at 75. 0 ft/s at 70
Example #8: A football is kicked upwards at 75.0 ft/s at 70.0o above the horizontal from the top of a 90.0 foot tall building.
(a) What is the maximum height of the ball above the ground?

51. Add this onto the 90.0 ft tall starting height:
(b) How long does it take for the ball to hit the ground?

53. (c) What is the horizontal range of the football?

54. Stop Tuesday.
Do Problems 20, 22, 29 independently. For problem 29 see page 62 for a useful formula.

55. Warm Up Example #9. 10 Minutes.

56. Example #9: A cannon fires a round at an angle of 65
Example #9: A cannon fires a round at an angle of 65.0°, and it is in the air for s. Find (a) the initial velocity of the projectile.

57. (b) What is the range of the projectile?

58. (c) What is the velocity of the projectile, as magnitude and direction, at 10.00 s?

59. vx = m/s
Use Pythagorean theorem: q
vy = – m/s v
Use inverse tangent:
below horizontal

60. Example #10: In making a record jump, the truck “Bigfoot” jumped 208 feet. If the truck left the ramp at 69.3 mph, and the landing ramp was identical in angle and height, determine the angle of the launch ramp.
Last 4 words spoken by a Redneck?
“Hey Y’all, Watch This!”

61. From example #7, part (g), use:

62. Example #11: It is the last play of the game and Troy is losing by 2 points to Sunny (cough) Hills. They decide to attempt a 55.0 yard field goal. The kicker kicks the ball straight at the 10.0 foot tall goal posts. If he kicks the ball at 52.5 mph and 40.0° above the horizontal, does Troy win?
First, determine the time for the ball to travel 55.0 yard = 165 feet. Then determine the height of the ball (hopefully) above the ground at this time. Is this height greater than 10.0 feet?

64. Alternatively….. See the solution to problem 29 in the textbook.

65. Example #12: Jürgen releases a shot-put 2
Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it?
Why can’t we use the range formula to find Vo?
But…. We could use formula that show the relationships between Y, X, Vo, g, and θ.

66. Example #12: Jürgen releases a shot-put 2
Example #12: Jürgen releases a shot-put 2.00 m above the ground at an angle of 45.0° above the horizontal. If his toss is 20.9 m, how fast did he release it?
Substitute Dx into Dy and solve for vo.

69. Warm UP to Test Tomorrow.
From memory, notes, text or homework write down all useful formulas
Homework Check
Sample Test
Problems 20, 22,29,30,31,32,35,
67, 75