1. Projectile Motion Unit 2.6
2 Dimensional Motion
Projectile Motion
Unit 2.6

2. Projectile motion
We have learned to “resolved” vectors into components. Now we will use the components to understand and predict the motion of objects thrown into the air.

3. Projectile motion
How can you know the displacement, velocity, and acceleration of a ball at any point in time during its flight?

4. Projectile motion
All of the kinematic equations from Unit1 could be rewritten in terms of vector quantities.

5. Projectile motion
When an object is propelled into the air in a direction other than up or down, the velocity, acceleration, and displacement of the object do not all point in the same direction.

6. Projectile motion
This makes the vector forms of the equations difficult to solve.

7. Projectile motion
By resolving vectors into components, you can apply the simpler one dimensional forms of the equations to determine the resultant vectors.

8. Projectile motion
Example: A long jumper in the air has both a horizontal and vertical component.

9. Projectile motion

10. Projectile motion
In this section, we will focus on the form of 2 dimensional motion called projectile motion. Projectile motion is free fall with an initial horizontal velocity.

11. Projectile motion
Objects that are thrown or launched into the air and are subject to gravity are called projectiles.

12. Projectile motion
The path of a projectile is a curve called a parabola.

13. Projectile motion
Many people mistakenly believe that projectiles eventually fall straight down, much like a cartoon character does after running off a cliff.

14. Projectile motion
However, if an object has initial horizontal velocity in any given time interval, there will be a horizontal motion throughout the flight of the projectile.

15. Projectile motion
We will assume that the horizontal velocity is constant. If we were accounting for air resistance, then velocity would not be constant since a projectile slows down as it collides with air particles.

16. Projectile motion
Hence, the true path of a projectile traveling through Earth’s atmosphere is not a parabola.

17. Projectile motion
To understand the motion a projectile undergoes, let’s examine the following:

18. Projectile motion
The red ball was dropped at the same instant the yellow ball was launched horizontally. If air resistance is disregarded, both balls hit the ground at the same time.

19. Projectile motion
By examining each ball’s position in relation to the horizontal lines and to one another, we see that the two balls fall at the same rate.

20. Projectile motion
The may seem impossible because one is given an interval velocity and the other begins from rest. But if the motion is analyzed one component at a time, it makes sense.

21. Projectile motion
First, consider the red ball that falls straight down. It has no motion in the horizontal direction. In the vertical direction, it starts from rest and proceeds in free fall.

22. Projectile Motion
Thus, the kinematics equation from Unit 1can be applied to analyze the vertical motion of the falling ball.

23. Projectile motion
Note that the acceleration , a, can be rewritten as –g because the only vertical component of acceleration is free-fall acceleration. Note also that Δy is negative.

24. Projectile motion
Vertical motion of a projectile that falls from rest:
Δ y = -1/2g (Δt)2
Horizontal motion of a projectile:
Δx = vxΔt

25. Projectile motion Δy = vi(sinθ)Δt -1/2g(Δt)2
Projectiles launched at an angle:
Δx= vi(cosθ)Δt
Δy = vi(sinθ)Δt -1/2g(Δt)2

26. Projectile motion
To find the velocity of a projectile at any point during its flight, find the vector sum of the components of the velocity at that point.

27. Projectile motion
Use the Pythagorean theorem to find the magnitude of the velocity, and use the tangent function to find the direction of the velocity.