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# Special Case of 2D Motion: Projectile Motion - PowerPoint PPT Presentation

Special Case of 2D Motion: Projectile Motion . Recall Gravity. The acceleration due to gravity What goes up. Must come down. a g = g = 9.8 m/s 2 [Down]. V f = 0 m/s. V i = 0 m/s. V i = 10 m/s. V f = -10 m/s. Concept Check.

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### Special Case of 2D Motion:Projectile Motion

• 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

• 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