1. Kinematics Projectile Motion

2. Objectives Define a projectile Describe the trajectory
State which variables remain constant during a projectile motion
Solve projectile motion problems
Sketch the kinematic graphs for projectile motion

3. Projectile
An object launched, thrown, or fired under the influence of gravity only.
The trajectory of a projectile is parabolic
What is the significance of “gravity only” ?
Gravity refers to the acceleration of gravity.
“Gravity only” implies that the acceleration of gravity is the only acceleration that projectiles experience.
The acceleration of gravity operates in the vertical direction only, which means that projectiles are accelerated in the y-direction.
There is NO horizontal acceleration due to gravity, which means that projectiles move at constant velocity in the x-direction.

4. Components of the Initial Launch Velocity
In most problems the launch velocity is given along with the angle of the launch. The angle of the initial launch is usually in the 1st or 4th quadrants.
The first step in most problems is determining the x- and y-components of the initial velocity vector.
Examples:
50 m/s, 37o above +x
50 m/s, 37o below +x
50 m/s horizontally
v0x = v0 cos 
v0x = 50 cos 37o
v0x = +40 m/s
v0y = v0 sin 
v0y = 50 sin 37o
v0y = +30 m/s
v0x = v0 cos 
v0x = 50 cos 37o
v0x = +40 m/s
v0y = v0 sin 
v0y = 50 sin 37o
v0y = 30 m/s
v0x = v0 cos 
v0x = 50 cos 0o
v0x = +50 m/s
v0y = v0 sin 
v0y = 50 sin 0o
v0y = m/s

5. Independence of Motion
Projectiles accelerate in the y-direction and move at constant velocity in the x-direction.
The mathematics of these two motions is completely different.
As before the motion in the x- and y-directions are mathematically independent.
You can solve in the x-direction using only x-variables and the constant velocity equation, while ignoring the y-direction completely.
You can also solve the y-direction using only y-variables and the kinematic equations for acceleration, while ignoring the x-direction.
While the x- and y-motions solve independently they take place in simultaneous time.
Time is the one variable shared by both the x- and y-directions. Once time is solved in one direction, it can then be applied to solve equations in the other direction. This makes time a critical variable.

6. Projectile Motion Variables
v0y= v0 sin θ θ0 v0
v0x= v0 cos θ
Most projectile motion problems begin by giving launch data.
v0 Magnitude of initial launch velocity.
θ0 Angle of the launch velocity.
When ever you encounter a vector at an angle you must find its components.
This is a critical step in many problems through out the entire year, and should be done automatically from now on.
v0x Component of the initial velocity in the x direction, v0x = v0 cos θ
v0y Component of the initial velocity in the y direction, v0y = v0 sin θ

7. Projectile Motion Variables
v0y= v0 sin θ
v0x= v0 cos θ θ0 v0 Δy Δx
The projectile then travels in a parabolic trajectory.
It may experience displacements in one or both directions (x and/or y).
Δx Range (change in horizontal position measured from the launch).
Δy Altitude (change in vertical position measured from the launch).
In a time
t Time of flight

8. Projectile Motion VariablesΔx Δy
v0y= v0 sin θ
v0x= v0 cos θ θ0 v0 vx θ vy v
The projectile finishes the problem with an overall final velocity.
v Magnitude of final velocity at the designated end of the problem.
θ Angle of the final velocity.
This is also a vector at an angle, and it splits into components
vx Component of final x direction velocity.
vy Component of final y direction velocity.

9. Modified for y direction Modified for x direction
Projectile Motion Equations Δx Δy
v0y= v0 sin θ
v0x= v0 cos θ θ0 v0 v θ vy vx
Original Equation
Modified for y direction
Modified for x direction
REMEMBER: Motion in the x direction is constant velocity, ax = 0 .

10. Modified for y direction Modified for x direction
Projectile Motion Equations
v0y= v0 sin θ
v0x= v0 cos θ θ0 v0 Δx Δy v θ vy vx
Result
y-direction:
Free fall
X-direction:
Constant velocity
Modified for y direction
Modified for x direction

11. + − + − Sign on the Variables Δy vy A B C D E
Identify the correct variable signs at key places along a projectiles trajectory. A
Moving upward B
At max height C
Moving downward, above initial position D
Moving downward, even with initial position E
Moving downward, below initial position Δy vy + − + −

12. Analyzing an Upward Launch
For this presentation g = 10 m/s2 . This simplifies example calculations.
x-direction
y-direction
Split into x and y adding subscripts
ax = 0 and ay = g
Horizontal displacement due to the x velocity.
Vertical displacement due to gravity g .
Vertical displacement due to the y velocity.

13. The graph at the left is a position- position graph ( y – x ).
Plot a projectile that is launched at m/s at an angle of 45o.
Step 0: Find components.
If there was no gravity the object would move at constant velocity in both the x and y directions.
It would move 20 m right and 20 m up every second resulting in the path described by the dashed line. 20
y (m) 40 60 80
100
−20
x (m)
−40 v0
v0x
v0y
What does gravity do?
Gravity only affects the y direction adding acceleration to the equation.

14. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

15. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

16. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

17. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

18. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

19. t 1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40

20. The dashed line represent the effect of velocity alone.1 2 3 4 5 20
y (m) 40 60 80
100
−20
x (m)
−40
The dashed line represent the effect of velocity alone.

21. The drop due to gravity always follows this pattern.1 2 3 4 5
−125 20
y (m) 40 60 80
100
−20
x (m)
−40
The drop due to gravity always follows this pattern.
−80
−45
−20 −5

22. Examine the Components of Velocity Each Second20
y (m) 40 60 80
100
−20
x (m)
+10 m/s
+20 m/s
0 m/s
+20 m/s
−10 m/s
−20 m/s
−30 m/s
x-direction is constant velocity.
In the y-direction gravity takes away 10 m/s (9.8) every second.

23. Facts Worth Knowing x-motion is always constant velocity and positive.20
y (m) 40 60 80
100
−20
x (m)
+10 m/s
+20 m/s
0 m/s
+20 m/s
−10 m/s
−20 m/s
x-motion is always constant velocity and positive.
Any two point with the same altitude
The speed (magnitude of velocity) is the same.
The direction is opposite: Positive on the way up and negative on the way down
At the very top there is no y-velocity
Velocity at max height equals the initial x-direction velocity, vmax height = v0x.

24. Solving Max Height Problems
Determining maximum height
HIDDEN ZERO: At maximum height the vertical velocity is zero, vy = 0 . While the projectile does have a horizontal velocity, this is irrelevant when solving independently in the y-direction.
The best equation to determine maximum height is:
Substituting zero for vy and rearranging for height results in:
Determining the speed or velocity at maximum height
This you should already know. The instantaneous velocity at max height is horizontal, and since velocity is constant in the x-direction the velocity and speed at max height are equal to the x-component of the initial velocity.

25. A projectile is launched at an angle of 37o above the horizontal with an initial speed of 50 m/s.
Example 1
When given the magnitude and direction of the initial launch determine the components of initial velocity first.
a. Determine the max height reached by the projectile.
b. Determine the time to reach max height.
c. Determine the speed of the projectile at max height.
v0y
v0x θ0 v0 Δx Δy v θ vy vx

26. Solving For Locations Other Than Maximum Height
Most problems begin by solving one of the following equations for time, and then substituting time into the other equation.
The equation on the left is solved exactly the same as it was in the free fall assignment, and the equation on the right is pretty simple.
Keep three things in mind:
Use the components of the initial velocity, and NOT the initial velocity.
The left equation can only contain y-direction variable
The right equation can only contain x-direction variable.
Use the other equations as necessary.

27. Example 2 v0x = 40 m/s , and v0y = 30 m/s
A projectile is launched at an angle of 37o above the horizontal with an initial speed of 50 m/s. During its flight the projectile achieves a range of 200 m.
Example 2
v0x = 40 m/s , and v0y = 30 m/s
a. Determine the time of flight.
b. Determine the height of the projectile relative to the launch point.
c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed
v0y
v0x θ0 v0 Δx Δy v θ vy vx

28. A projectile is launched at an angle of 37o above the horizontal with an initial speed of 50 m/s. The projectile’s time of flight is 5 seconds.
Example 3
v0y
v0x θ0 v0 Δx Δy v θ vy vx
a. Determine the range of the projectile.
b. Determine the height of the projectile relative to the launch point.
c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed
The solution to this part solved the same as Example 1

29. A projectile is launched at an angle of 37o above the horizontal with an initial speed of 50 m/s. The projectile lands 25 m above its initial launch height.
Example 4
v0y
v0x θ0 v0 Δx Δy v θ vy vx
a. Determine the time of flight.
b. Determine the projectiles range.
c. Determine the speed of the projectile. NOTE: This refers to the diagonal speed
See the two previous examples. This calculates the same each time.

30. Landing at the starting height
Two Special Launches
There are two frequently encountered launches, that have hidden zero quantities. Hidden zeros often means short cuts and easier solutions.
Landing at the starting height
Horizontal Launches
Hidden zero: y = 0
Hidden zero: v0y = 0

31. Return to Initial Height
Landing at the Initial Height 20
y (m) 40 60 80
100
−20
x (m)
vtop = v0x
+20 m/s
0 m/s
+20 m/s v0
v = v0 θ −θ
−20 m/s
Maximum Altitude
Return to Initial Height
Other Info
2nd half look like a horizontal launch.

32. Landing at the Initial Height
There is a pattern associated with launch angles for objects returning to their initial height.
15o
75o
30o
60o
45o
Any two angles of launch that add up to 90o will arrive at the same landing position.
The greatest range is achieved by launching at an angle of 45o.
Maximum height is achieved by launching at 90o.

33. A projectile is launched at 141.42 m/s to achieve maximum range.
Example 5
When given launch velocity and launch angle, determine launch components.
HIDDEN VALUE: Maximum range means the launch angle is 45o.
a. Determine the projectile’s maximum height.
HIDDEN ZERO: At max height vy = 0 .
b. Determine the time to reach maximum height.
c. Determine the projectile’s speed at maximum height.

34. A projectile is launched at 141.42 m/s to achieve maximum range.
Example 5
d. Determine the total time of flight to landing.
HIDDEN ZERO: There is no mention of landing higher or lower, y = 0 .
e. Determine the projectile’s maximum range.
This is 4 times the maximum height. Is this always true in max range problem?
f. Determine the projectile’s landing speed.

35. A ball is thrown horizontally with a speed of 25 m/s from the top of a 45 m tall building.
Example 5
When given launch velocity and launch angle, determine launch components.
HIDDEN ZERO: In horizontal launches v0y = 0 .
a. Determine the time of flight.
Just as in free fall problems, the vertical displacement is negative. This means the minus signs will cancel. Everyone knows this, and for dropped objects in free fall and horizontally launched projectiles, most people ignore the minus signs.
b. Determine the projectile’s range.

36. Graphing a Horizontal Launch
x (m)
y (m) 20 40 60 80
t (s) 1 2 3 4
Clearly x and y solved independently, yet the events took place in simultaneous time

37. Projectile Motion Graphs
Position Graphs: These look exactly like the flight path. x y
Horizontal Launch
Upward Launch
Acceleration Graphs: There is only one acceleration, gravity = −9.8 t g
Horizontal Launch
Upward Launch

38. Projectile Motion Graphs
Horizontal Velocity Graphs: Constant Velocity (no acceleration in x ) t vx
Horizontal Launch
Upward Launch
Vertical Velocity Graphs: a is the rate of Δv (subtract 9.8 each second) t vy
Horizontal Launch
Starts at rest vertically
Upward Launch
Thrown up (+vy) initially