1. What is projectile motion? The only force acting on the objects above is the force of the Earth.

2. With no Force of the Earth….

3. What do we actually see?

4. What do you notice about the horizontal and vertical components of the motion?

5. What if we launch at an angle rather than horizontally?

6. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane? a.below the plane and behind it. b. directly below the plane c. below the plane and ahead of it

7. What do you notice about the values of Vx and Vy?

8. What do you notice about Vx and Vy when we launch at an angle?

9. Kinematics Equations for Projectile Motion: Y = ½ gt 2 (This is vertical displacement of the projectile when starting from rest) X = v 0x t (This is horizontal displacement of the projectile) V fx = V 0x Why is this true? V fy = V 0y + a y t Y = v 0y t + ½gt 2 if there is an initial velocity in the y direction.

10. How do we solve these problems? The vertical motion of a projectile is just like that of an object dropped or thrown straight up or straight down. The force of the Earth is acting on the object, so a = g (-9.8 m/s²). The horizontal motion of a projectile is just like any constant velocity problem. No horizontal forces are acting on the object so Vₒₓ is the only velocity and aₓ = 0.

11. Now you try… Fill in the table below indicating the value of the horizontal and vertical components of velocity and acceleration for a projectile.

12. The answer: Time (s)Vx (m/s)Vy (m/s)Ax (m/s/s)Ay (m/s/s) 015.029.40-9.8 115.019.60-9.8 215.09.80-9.8 315.000-9.8 415.0-9.80 515.0-19.60-9.8 615.0-29.40-9.8

13. The diagram to the right shows the trajectory for a projectile launched non- horizontally from an elevated position on top of a cliff. The initial horizontal and vertical components of the velocity are 8 m/s and 19.6 m/s respectively. Positions of the object at 1-second intervals are shown. Determine the horizontal and vertical velocities at each instant shown in the diagram.

14. The v x values will remain 8 m/s for the entire 6 seconds. The v y values will be changing by 9.8 m/s each second. Thus, v y =9.8 m/s (t = 1 s) v y = 0 m/s (t = 2 s) v y = -9.8 m/s (t = 3 s) v y = -19.6 m/s (t = 4 s) v y = -29.4 m/s (t = 5 s) v y = -39.2 m/s (t = 6 s)

15. How do we find the initial component velocities if we are given the angular velocity? Vₓ 0 = V cos 60° = V x V y0 = V sin 60° Notice that V x 2 + V y 2 = V 2

16. What is the maximum height the ball reaches? Ball reaches maximum height when V y = 0, so V y = V y0 + a y t becomes 0 = 43 m/s +(-9.8)t -43 = -9.8t t = 4.4 s; Y = ½a y t 2 so Y = 94.3m Ball’s total hang time is 2 x the time it takes to reach max height, so total time = 8.8 s. How far does the ball travel horizontally in the time it is in the air?

17. Remember motion in the x direction is independent of motion in the y direction. X = X 0 + V x0 t + ½ a x t 2 X 0 = 0 and a x = 0, so X = V x0 t = (25 m/s)(8.8s)= 220m. So the projectile travels up and down with a max height of 94.3 meters. When it lands back on the ground, it is 220 m from its starting position.

18. Practice Problem: You accidentally throw your car keys horizontally at 8.0 m/s from a cliff 64-m high. How far from the base of the cliff should you look for your keys?

19. The answer: V 0x = 8.0m/s 2,a x = 0, V 0 y = 0, a y = -9.8 m/s 2 Y f = -64 m - 64m = ½ (-9.8 m/s 2 )t 2, t = 3.6 s In that 3.6 seconds, how far do the keys travel horizontally? X = V 0x t=8.0m/s 2 (3.6s) = 28.8m