1. Projectile Motion Two kinds of motion in one Constant motion Accelerated motion

2. The path of a projectile is a PARABOLA

3. The path of a projectile is shown above. Assume equal amounts of time have passed at each of the points on the projectiles path above. Question - Has the projectile traveled equal amounts of distance in equal amounts of time horizontally? How about vertically?

4. Answer - There is equal horizontal spacing in equal amounts of time on the picture indicating that horizontally the object has CONSTANT MOTION. However, vertically there is unequal spacing indicating that the object is either losing velocity or gaining velocity. Therefore vertically the object has ACCELERATED MOTION.

5. The arrows drawn below are vectors that represent the objects horizontal and vertical velocity. Notice the vertical ones change size and the horizontal velocity vectors stay the same size.

6. Why represent velocity with vector arrows?

7. Look at these examples v=10 m/s East v=20 m/s East Notice that both vectors are pointing to the right (East). However, one vector arrow is twice as long as the other because the magnitude of this velocity is twice as much as the other.

8. Other Examples v i =25m/s at 65° above horizontal v i =12m/s at 30°below horizontal

9. So Why use arrows to represent vectors? Arrows point, and vectors need a direction. The direction of the arrow is the direction of the vector Arrows can be drawn with different lengths. The length of the arrow is drawn proportional to the size of the vector quantity.

10. Now, projectile motion is both kinds of motion, constant and accelerated motion. For calculation purposes these two types of motion must be considered separately. Any given velocity or displacement (vectors) in a projectile motion problem must be broken down into its horizontal and vertical components.

11. An object is launched with an initial velocity of 20 m/s at a 35° angle up from horizontal. How much of this velocity is vertical and how much is horizontal? 35° V i =20m/s Example

12. An object is launched with an initial velocity of 20 m/s at a 35° angle up from horizontal. How much of this velocity is vertical and how much is horizontal? V iy =? V i =20m/s V x =? 35º

13. V i =20m/s V x =? 35º V iy =?

15. Equations Horizontal - Constant Motion Vertical - Accelerated motion

16. The airplane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050m. Determine the time required for the care package to hit the ground. Example one - projectile horizontally launched

17. V i =v x =115m/s Note: since this projectile was launched horizontally, there is no need to “break it down” into x and y. The velocity is all horizontal and no vertical.

18. How far horizontally does the care package fly? Givens V x =115m/s ∆x=? t=14.6s(same as vertical t)

19. Example 2 - a projectile launched at an angle A football is kicked at 22m/s at an angle of 40 degrees. Ignoring air resistance, how long will the football be in the air if it lands at the same height it was kicked?

20. Vi=22m/s 40° V iy =? V i =22m/s V x =? 40º Example 2 - projectile launched at an angle

21. V i =22m/s V x =? 40º V iy =?

22. Write the givens v x =16.9m/s ∆x=? t=? v iy =14.1m/s t=? g=-9.8m/s/s ∆y=0m (lands at same vertical height) Remember you can try to find time using the horizontal info or the vertical info. In this case, choose a vertical equation.

23. Plugging in the numbers, and leaving off the units for the sake of looks….. It takes 2.9 seconds to hit the ground

24. Vi=22m/s 40° ∆y=? In other words, what is the ∆y half way through the flight? Givens v iy =14.1m/s g=-9.8m/s/s V fy =0m/s ∆y=? There is no vertical velocity at the highpoint

25. Givens v iy =14.1m/s g=-9.8m/s/s V fy =0m/s ∆y=? Using a different vertical motion equation…

26. Vi=22m/s 40° ∆x=? Givens v x =16.9m/s t=2.9s ∆x=? Using…