Special Case of 2D Motion: Projectile Motion

1. Special Case of 2D Motion:Projectile Motion

2. 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

3. Concept Check • What determines how long it will take an object to reach the ground when released with an initial velocity of zero?

4. Projectile motion • 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

5. 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

6. Solution **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

7. Still confused, look at Example 1 on pg 77