1. 2-Dimensional Kinematics Unit 2 Presentation 2

2. Projectile Problems  Projectile Motion: The two- dimensional (x and y) motion of an object through space  RULE: The horizontal (x) and vertical (y) motions of an object are completely independent of each other, except for the time taken.

3. Types of Projectile Motion  Type I Projectile Motion Height h Distance d x Initial Velocity v o Note that all of the initial velocity is in the x direction. There is NO initial velocity in the y direction. Hence, this motion is free-fall.

4. Types of Projectile Motion  Type II Projectile Motion Note the symmetry between the two halves of motion. This is an important aspect of Type II Projectile problems.  Distance d x Height h Initial Velocity V o

5. More Type II Projectiles Consider the velocity of the object at the point of maximum height.  Distance d x Height h Initial Velocity V o V max is only in the y direction

6. Types of Projectile Motion  Type III Projectile Motion Initial Height h o Distance d x Initial Velocity v o   Maximum Height h max

7. Type I Example Problem  A bowling ball rolls off a table 5.0 m high with an initial velocity of 8.0 m/s. (a) What is the horizontal range of the bowling ball (i.e., how far away from the table does it strike the floor)? (b) What are the horizontal and vertical components of the velocity immediately before impact?

8. Type I Example Problem (cntd)  First, draw a picture. 5.0 m Distance d x 8.0 m/s

9. Type I Example Problem (cntd)  Now, consider that the ball is in free-fall motion. Find time. Now, find horizontal distance.

10. Type I Example Problem (cntd)  Part (b) Is there any acceleration in the x direction? NO Hence, the final x velocity is the same as the initial x velocity (8.0 m/s)  Now, find the y component of the final velocity What is the acceleration in the y-direction?  g = -9.8 m/s 2, v o = 0 m/s, t = 1.01 sec Hence, the final velocity is

11. Type II Example Problem A golfer hits a golf ball at an angle of 25° above the horizontal at an initial velocity of 20 m/s. (a) How long does it take the golf ball to reach its maximum height? (b) What is the golf ball’s maximum height? (c) How far does the golf ball go? ° Distance d x Height h 20 m/s

12. Type II Example Problem  First, break down the initial velocity vector into its components. V x = V o cos  = 20 m/s cos (25°) = 18.13 m/s V y = V o cos  = 20 m/s sin (25°) = 8.45 m/s Now, consider the y-velocity at the maximum height. It is zero. The only acceleration in the y-direction is the acceleration due to gravity. Hence:

13. Type II Example Problem  Now, lets find the maximum height: Or, alternatively:

14. Type II Example Problem  Finally, lets find the horizontal range:

15. Type III Example Problem  Now, lets combine the previous two problems, hitting the golf ball off of a table with the same values. 5 m Distance d x 20 m/s  ° Maximum Height h max Lets find: 1)Time until it strikes the ground 2)Horizontal Distance 3)Maximum Height 4)Components of the Final Velocity

16. Type III Projectile Problem  First, we need to break down the initial velocity vector into its components: Type III projectiles are NOT in free fall! Use the big equation:

17. Type III Projectile Problem  How do we solve ?  Use the Quadratic Formula! x = 2.19 sec or x = -0.46 sec Only one of these two answers is the correct answer. Obviously, you cannot have negative time. Hence, the time taken to strike the ground is 2.19 sec.

18. Type III Projectile Problem  Now, lets find the horizontal range of the projectile. Remember, there is no acceleration in the x direction.

19. Type III Projectile Problem  Now, lets find the maximum height. Remember, the y component of the velocity is zero at the maximum height. Adding 3.64 m to our original height of 5.0 m gives us a total maximum height of 8.64 m.

20. Type III Projectile Problem  Now, lets find the components of the final velocity Remember, there is no acceleration in the x-direction, so the final x-component is equal to the initial x- component (18.13 m/s)  Lets find the y-component of the final velocity: Hence, the components of our final velocity are: