1. Projectile Motion
Physics Honors

2. The Motion Formulas
Good News/Bad News: These are the same formulas we used for linear motion. Do you know them?
If the answer is “NO”, then memorize them NOW!

3. Things we know to be true about all projectiles.
We assume NO AIR RESISTANCE! (Welcome to “Newtonia”), therefore…
The path of a projectile is a parabola.
Horizontal motion is constant velocity.
Vertical motion is in “free-fall”.
Vertical velocity at the top of the path is zero
Time is the same for both horizontal and vertical motions.

4. How are the formulas different for projectiles
How are the formulas different for projectiles? They must be applied along only one axis at a time.
horizontal or “x” – direction
vertical or “y” – direction
Remember that for projectiles, the horizontal and vertical motions must be separated and analyzed independently. Remember that “ax” is zero and “ay” is acceleration due to gravity “g = 9.81 m/s2”.

5. Projectiles Launched Horizontally

6. A typical physics problem
A cannon ball is shot horizontally from a cliff. vx
What do we know?
For all projectiles…
height dy
Range, dx
Hint: You should always list your known values at the beginning of any problem and assign those values variables.

7. Where do we start? A cannon ball is shot horizontally from a cliff. vx
Knowns:
height dy
Range, dx
Remember to keep the horizontal and vertical motions separate. Time is the factor that ties them together. Since the vertical motion is treated like “free-fall” and we have more info about the vertical, we should use that to find time first.

8. Let’s put numbers on it. What do we want to find?
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m
height
Range, dx
Add the given values to our list of known values. Now that the diagram is drawn and labeled and we have identified and listed all of our “known” and “given” values for the problem, let’s begin by finding time.

9. Finding time of flight for a horizontally launched projectile.
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m
height
Range, dx
Since we know more values for vertical motion, let’s use it to find time. Start with the distance equation…
Now solve for t…

10. Finding range (horizontal distance)
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m
height
Calculated:
Range, dx
Now that we know time, let’s find dx. Remember that horizontal motion is constant. Let’s use the distance formula again. This time in the x – direction…

11. Finding final velocity (magnitude and direction)
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m Vf
height
Calculated:
Range, dx
Final velocity requires a little more thought. Remember that velocity is a vector quantity so we must state our answer as a magnitude (speed that the projectile strikes the ground) and direction (angle the projectile strikes the ground). Also remember horizontal velocity is constant, therefore the projectile will never strike the ground exactly at 90°. That means we need to look at the horizontal (x) and vertical (y) components that make up the final velocity.

12. Finding final velocity (magnitude and direction)
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m Vf
height
Calculated:
Range, dx θ
Let’s look more closely at the vector, vf. To help see it better, let’s exaggerate the angle. Since x- and y- motion are separate, there must be components.
So, we have the x-component already due to the fact that horizontal velocity is constant. Before we can find vf, we must find the vertical component, vfy. Vf
Vfy θ
Vfx = 5 m/s

13. Finding final velocity (magnitude and direction)
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m Vf
height
Calculated:
Range, dx θ
To find vfy, remember that vertical motion is in “free-fall” so it is accelerated by gravity from zero to some value just before it hits the ground.
Calculating vfy: Vf
Vfy θ
Vfx = 5 m/s
Still not finished. Gotta put components together for final.

14. Finding final velocity (magnitude and direction)
A cannon ball is shot horizontally at 5 m/s from a cliff that is 35 m high. Find time of flight, t = ? Find range, dx = ? Find final velocity, vf = ?
Vx= 5 m/s
Knowns:
Givens:
dy=35 m Vf
height
Calculated:
Range, dx θ
Now we know both the x- and y- components of the final velocity vector. We need to put them together for magnitude and direction of final velocity.
Putting it together to calculate vf:
Vfy=26.2 m/s Vf θ
Vfx = 5 m/s
Final Velocity = 26.7 m/s, 79.2°

15. Projectiles Launched at an Angle

16. Separating the horizontal and vertical motions.vi θ
viy
vix
Since the initial velocity represents motion in both the horizontal (x) and vertical (y) directions at the same time, we cannot use it in any of our equations. Remember, the most important thing about projectiles is that we must treat the horizontal (x) and vertical (y) completely separate from each other. So…we need to separate “vi” into its x- and y- components. We will use the method we used for vectors.
Now that we have the components of the initial velocity, we will use only those for calculations. **Never use the original velocity at the angle in an equation! vi
viy θ
vix

17. Symmetry of the projectiles path.vf θ
vfx vi θ
viy
vfy
vix
A projectile’s path is a parabola, ALWAYS. That means if a projectile is launched and lands at the same height, there will be symmetry. The angle of launch and angle of landing will be equal. The initial velocity and the final velocity will be the same magnitude. Also, that means the components will be the same.
Since the vertical motion is the same as a ball that is thrown straight up or dropped straight down (in free-fall), the y-components are equal and opposite. vf θ
vfx
vfy
Since the horizontal motion is always at constant velocity…

18. What do we know? Step 1: List known values! Draw and label picture.
Vy top = 0 vi θ
Viy
dymax =height Vx
Range, dx
Knowns (for all projectiles):

19. Keeping the horizontal and vertical motions separate!
Step 1: List known values! Draw and label picture.
Vy top = 0 Vy
dymax =height vi θ Vx
Range, dx
Knowns (for all projectiles):
Step 2: Divide initial velocity into horizontal (x) and vertical (y) components.
Step 3: Find time if possible. Use vertical motion.

20. Finding time for projectiles that start and end at the same height
Finding time for projectiles that start and end at the same height. (Using distance formula.)
Almost every projectile problem can be solved by starting with the displacement equation to solve for time. In this case…
Vy top = 0 Vy
dymax =height vi θ
Note: There are three ways to find time for this problem. You may use any of them you wish.
Range, dx
Finding time – Method 1:
Since the initial and final vertical positions are both the same, vertical displacement dy = 0.
Now solve for time. Yes, it is a quadratic equation! This will be the time for the entire flight.
NOTE: If you want to find maximum height you will only use half the time. If you want to find range, use the total time.

21. Finding time for projectiles that start and end at the same height
Finding time for projectiles that start and end at the same height. (using symmetry)
Vy top = 0 vf θ vi θ Vy
vfy
dymax =height Vx
vfx
Range, dx
Find time – Method 2:
Use vertical motion and symmetry. Remember that the y component of initial and final velocities are equal and opposite.

22. Projectiles that land higher than they startvo y θ x
Vertical displacement y-y0 = y.
Now solve for time, this will be the time for the entire flight. Then you can use that time to find the horizontal distance at which it was at that height.

23. Projectiles that are thrown downward
Vertical displacement is “-y”
Now solve for time, this will be the time for the entire flight. Then you can use that time to find the horizontal distance at which it was at that height. vo θ -y x

24. Projectiles that land lower than they start.
Vertical displacement is “-y”
Now solve for time, this will be the time for the entire flight. Then you can use that time to find the horizontal distance at which it was at that height. vo θ -y x 24