1. Part I: Projectile Motion Part II: Circular Motion
2-Dimensional Motion
Part I: Projectile Motion Part II: Circular Motion

2. First of All, What is 2-D Motion?
Before, we talked about motion in one dimension (just the x axis)
Now we are investigating things that can move it 2 dimensions (x and y)! y x
2D Motion x
1D Motion

3. 2-D Motion Part I: Projectile Motion
A “projectile” is an object on which the only force acting is gravity.
These are things that fly through the air! (Examples: footballs, baseballs, cars flying off cliffs, bullets, cannon shells, and batman)
Usually they are thrown/fired/etc. but are not things like rockets, which still have something “pushing” them as they fly through the air!
As always, gravity will accelerate the object down.

4. The Dimensions are Independent
Since 2D motion is complicated, it is helpful to think of the x and y dimensions separately when we analyze motion.
This is because the x and y dimensions are completely independent.
This lets us use all of our favorite 1D equations when solving problems!

5. Notice that after this idiot jumps off the cliff his velocity in the x dimension does not change.
BUT……His velocity in the y dimension is increasing due to gravity.
The same is true for this cannonball:

6. Gravity What causes projectiles to accelerate?
In which dimension does gravity exist?
The vertical (y)!
The acceleration is always 9.81 m/s2 downward. This is called “acceleration due to gravity” or “g”
Since acceleration only exists in the y dimension this means it is always zero for the x dimension.

7. Horizontal Velocity
Since we have established that acceleration only exists in the y dimension and the acceleration in the x dimension is zero, what does this mean about the velocity of a projectile in the x dimension?
It is always constant!
This needs to be remembered (by YOU)!

8. For an object fired at the same velocity in all cases, with which angle does the projectile go farthest?

9. For an object fired at the same velocity in all cases, with which angle does the projectile go highest?

10. For an object fired at the same velocity in all cases, with which angle does the projectile spend the most time in the air?

11. So all that means…
45o Is the angle at which a projectile will go the farthest. (best range)
The higher the object goes up, the longer it is in the air. (more height = more time)
The greater the angle, the greater the hang time.
If the angle stays the same and you increase velocity, you will also increase hang time.

12. POP QUIZ
Is this even possible?!?! 1 2 3 4

13. If you fire a rifle horizontally and drop a bullet at the same time and height which will hit the ground first.
The fired bullet
The dropped bullet
Both will drop at the same time
[Default]
[MC Any]
[MC All]

14. Solving Problems
We are going to take some time to go over the problem solving process for these projectile motion problems.
It is a “modified” version of GUESS.
The best part is that you don’t have to learn any complicated equations… we will use the ones you already know from the last unit! It’s easy!

15. Givens x y vo = v = ax = t = x = ay = y =+ a v d
Of course, the first step to solving any problem is to list your givens.
Always begin by drawing a diagram representing the motion. And remember to include your arrows for initial velocity, acceleration, displacement, and the positive direction.
Remember, we are separating the variables into the two dimensions.
You will write out the same givens table as before, but this time it will be a different set for each dimension (x or y)! x y
vo =
v =
ax =
t =
x =
ay =
y =

16. Equations The equations you will use are the same ones.
You will just be using the variables from one dimension (x or y) at a time.

17. Time
In projectile motion, we will treat the x and y variables as completely different sets.
This means the givens we use for one set cannot be used with the other!
However, Time is a variable that is always the same for both x and y. x y
vo = 25 m/s
v = 25 m/s
a = 0 m/s2
t = 4.0 s
x = 190 m
vo = 0 m/s
v = 48 m/s
ay = 10.0m/s2
y = 22 m

18. Sample Problem x y vo = v = a = t = x = ay = y = 95m/s 0m/s ? 95m/s
A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land?
As always, we will start by drawing a diagram of the situation. Remember to label any givens and include a direction vector for v, d, a, and +
After you have drawn your picture, fill in your Givens table.
And of course our unknown variable is x. Always circle the unknown in the table. +
Vx=95m/s x y
vo =
v =
a =
t =
x =
ay =
y =
95m/s
0m/s vy ay ?
95m/s
y = 60 m
0m/s2
10.0m/s2 + ? ? ?
60 m
x = ?? m

19. Sample Problem x y vo = v = a = t = x = ay = y = 95m/s 0m/s2 ? 0m/s
A cannon is placed at the edge of a cliff so the barrel is 60 meters above the ground. The cannon fires a projectile horizontally with a velocity of 95 m/s. How far from the base of the cliff does the projectile land?
dy= 60 m
dx = ?? m vy
Vx=95m/s ay +
Now determine the equations you will need to solve the problem.
Remember, we will keep the sets of variables completely separate (except time)!!!
x = vot + ½ at2
x = (95)(3.5) + ½ (0)(3.5)2
x = m
y = vot + ½ at2
t = √(2y / a)
t = √[(2)(60) / (10.0)] = 3.5 s x y
vo =
v =
a =
t =
x =
ay =
y =
This is the only equation we can use to solve for x. But we are missing time! We need to use the variables in the y-dimension to find time first.
95m/s
0m/s2 ?
0m/s
10.0m/s2
3.5 s
3.5 s ?
60 m

20. Practice Problem
Thelma and Louise drive their car off the edge of the grand canyon at a speed of 30 m/s and die 3.4 seconds later.
(a) how tall is the canyon at that spot?
Answer:
57.8 m
(b) how far away from the bottom of the canyon do they land?
102 m

21. Which of the following does not effect the hang time of a projectile?
Angled fired
Vertical velocity
Horizontal velocity
Height it was fired at.
[Default]
[MC Any]
[MC All]

22. Angled Projectile Motion
Occasionally, you will encounter problems where the projectile is initially fired at an angle.
This means that the initial velocity will have a horizontal and a vertical component.
You will need to calculate these components in order to fill in your givens table on each problem.

23. Vector Components
In order to find the components of a vector (like velocity) you will need to use those timeless Trigonometric Functions.

24. Vector Components
So we have a projectile that is fired at an angle of 30o with respect to the horizontal at a velocity of 50 m/s.
We need to think of it like the projectile is fired vertically and horizontally at the same time, giving it both components. vy
V=50m/s vx
Θ=30o

25. Vector Components
In order to calculate the components, we need to shift vy to make a right triangle.
Then we can use trig functions to solve for vy and vx like they are sides of a right triangle.
To solve for vx, we will use cosine because it is adjacent and we have the hypotenuse. vy
V=50m/s vx
Θ=30o

26. Vector Components
To solve for vx, we will use cosine because it is the adjacent side and we have the hypotenuse.
To solve for vy, use the same process but with sine. vy
V=50m/s vx
Θ=30o

27. Sample Problem x y vi = vf = ax = t = dx = ay = dy = ? ? ? ? 0 m/s2
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of 20m/s. What is the range of the soccer ball? (how far away does it land)
As always, start by drawing a
diagram, including all vectors. x y
vi =
vf =
ax =
t =
dx =
ay =
dy =
We need to use trig functions to
solve for vix and viy
And as always circle your
unknown variable.
6.84m/s ? ?
+18.8m/s
Begin to fill in what you can
on your givens table
6.84m/s ? ?
-18.8m/s
0 m/s2
-10.0 m/s2 ? ? ?
0 m a +
vix = VcosΘ
vix=(20)cos(70)
vix=6.84 m/s
V=20m/s vy
viy = VsinΘ
viy=(20)sin(70)
viy=18.8 m/s
Θ=70o vx dx +

28. Sample Problem x y vi = 6.84 m/s vf = 6.84 m/s ax = 0 m/s2 t = ?
A soccer ball is kicked at an angle of 70o from the horizontal with a velocity of 20.0m/s. What is the range of the soccer ball? (how far away does it land) x y
vi = 6.84 m/s
vf = 6.84 m/s
ax = 0 m/s2
t = ?
dx = ?
vi = m/s
vf = m/s
ay = -10.0m/s2
dy = 0 m
dx = vixt + ½ at2
dx = (6.84)(t) + ½ (0)t2
dx = (6.84)(3.76) + 0
dx = 25.7 m
vf = vi + at
t = (vf – vi) / a
t = (-18.8 – 18.8) / (-10.0)
t = 3.76 s
V=20m/s vx vy
Θ=70o + a dx