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Special Case of 2D Motion: Projectile Motion

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Special Case of 2D Motion:Projectile Motion

Recall Gravity

- The acceleration due to gravity
- What goes up.
- Must come down.

ag= g = 9.8 m/s2 [Down]

Vf = 0 m/s

Vi = 0 m/s

Vi = 10 m/s

Vf = -10 m/s

- What determines how long it will take an object to reach the ground when released with an initial velocity of zero?

- Motion of an object only under the influence of gravity (Path is a 2D curve)
- Can be broken up into two parts and analyzed separetely!

A

B

C

+

=

Non - uniform

uniform

Sample Problem

- A cannon on top of a 10 m wall fires a cannonball at an initial velocity of 10 m/s in the horizontal direction. How far will the cannonball go before it hits the ground?

C

G:

R: Δdx = ?

- In the “X” Direction
- vx = 10 m/s [R]
- Use Δdx = vx × Δt

X-Y Position

12

10

- In the “Y” Direction
- viy = 0 m/s
- diy = 10 m [D]
- ay = 9.8 m/s2 [D]
- Use Δdy = viyΔt +1/2 ay (Δt)2

8

6

4

2

0

2

4

6

8

10

12

14

0

**Choose +/- directions

[R] = +

[D] = -

Let’s work with the y-direction first

- In the “Y” Direction
- viy = 0 m/s
- diy = 10 m [D] = - 10m
- ay = 9.8 m/s2 [D] = - 9.8 m/s2
- Use Δdy = viyΔt +1/2 ay (Δt)2

-10 = 0 + ½ (-9.8) (Δt)2

-10 = ½ (-9.8) (Δt)2

2(10)/9.8 = (Δt)2

2.04 = (Δt)2

1.4 = Δt

Now that we have time we can go back to the x-direction

- In the “X” Direction
- vx = 10 m/s [R] = + 10 m/s
- Use Δdx = vx × Δt

Δdx = 10 m/s (1.4 s)

Δdx= 14 m

Therefore object will land 14 m away from the edge of the wall

- Still confused, look at Example 1 on pg 77