Projectile Motion or 2D Kinematics

1.  Projectile Motion or 2D Kinematics By Sandrine Colson-Inam, Ph.D

2. Outline What is a projectile Characteristics of a projectile's motion Horizontal and vertical components of velocity and displacement Initial velocity components Examples of problems

3. What is a projectile? A projectile is an object upon which the only force acting is gravity. A projectile is any object which once projected or dropped continues in motion by its own inertia and is influenced only by the downward force of gravity.

4. Projectile Motion and Inertia

5. Horizontal and Vertical Velocities A projectile is any object upon which the only force is gravity, Projectiles travel with a parabolic trajectory due to the influence of gravity, There are no horizontal forces acting upon projectiles and thus no horizontal acceleration, The horizontal velocity of a projectile is constant (a never changing value), There is a vertical acceleration caused by gravity; its value is 9.8 m/s/s, down, The vertical velocity of a projectile changes by 9.8 m/s each second, The horizontal motion of a projectile is independent of its vertical motion.

6. Vector diagrams for projectile motion

7. Horizontal and vertical displacement � Horizontally Launched Projectile

8. Displacement diagram of projectile motion Horizontal

9. Horizontal and vertical displacement : Non - Horizontally Launched Projectile

10. Check your understanding

11. Answers 1. It will take 4 seconds to fall 78.4 meters Use the equation y = 0.5 � g � t2 and substitute -9.8 m/s/s for g. The vertical displacement must then be subtracted from the initial height of 78. 4 m. At t = 1 s, y = 4.9 m (down) so height is 73.5 m (78.4 m - 4.9 m )? At t = 2 s, y = 19.6 m (down) so height is 58.8 m (78.4 m - 19.6 m )? At t = 3 s, y = 44.1 m (down) so height is 34.3 m (78.4 m - 45 m)? At t = 4 s, y = 78.4 m (down) so height is 0 m (78.4 m - 78.4 m)? 2. It will take 4 seconds to fall 78.4 meters Use the equation y = 0.5 � g � t2 and substitute -9.8 m/s/s for g. The vertical displacement must then be subtracted from the initial height of 78. 4 m. At t = 1 s, y = 4.9 m (down) so height is 73.5 m (78.4 m - 4.9 m )? At t = 2 s, y = 19.6 m (down) so height is 58.8 m (78.4 m - 19.6 m )? At t = 3 s, y = 44.1 m (down) so height is 34.3 m (78.4 m - 45 m)? At t = 4 s, y = 78.4 m (down) so height is 0 m (78.4 m - 78.4 m)?

12. Check your understanding

13. Answers 3.The vx values will remain constant at 15.0 m/s for the entire 6 seconds; the ax values will be 0 m/s/s for the entire 6 seconds. The vy values will be changing by -9.8 m/s each second. Thus, vy = 29.4 m/s (t = 0 s) vy = 19.6 m/s (t = 1 s)? vy = 9.8 m/s (t = 2 s) vy = 0 m/s (t = 3 s)? vy = -9.8 m/s (t = 4 s) vy = -19.6 m/s (t = 5 s)? vy = -29.4 m/s (t = 6 s)? The ay values will be -9.8 m/s/s for the entire 6 seconds. 4.The vx values will remain 8 m/s for the entire 6 seconds. The vy values will be changing by 9.8 m/s each second. Thus, vy =9.8 m/s (t = 1 s) vy = 0 m/s (t = 2 s)? vy = -9.8 m/s (t = 3 s) vy = -19.6 m/s (t = 4 s)? vy = -29.4 m/s (t = 5 s) vy = -39.2 m/s (t = 6 s)?

14. Initial components of velocity

15. Evaluating various info Determination of the Time of Flight Determination of Horizontal Displacement x = vix � t Determination of the Peak Height y = viy � t + 0.5 � g � t2

16. Equations of motion Horizontal motion Vertical Motion

17. Solving Projectile Motion Problems The following procedure summarizes the above problem-solving approach. Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side. Identify the unknown quantity which the problem requests you to solve for. Select either a horizontal or vertical equation to solve for the time of flight of the projectile. With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

18. Check your understanding A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal displacement, and the peak height of the football.

19. Answer State the problem Use the appropriate equations of motion

20. The Problem-Solving Approach The following procedure summarizes the above problem-solving approach. Use the given values of the initial velocity (the magnitude and the angle) to determine the horizontal and vertical components of the velocity (vix and viy). Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side. Identify the unknown quantity which the problem requests you to solve for. Select either a horizontal or vertical equation to solve for the time of flight of the projectile. For non-horizontally launched projectiles, the second equation listed among the vertical equations (vfy = viy + ay*t) is usually the most useful equation. With the time determined, use a horizontal equation (usually x = vix*t + 0.5*ax*t2 ) to determine the horizontal displacement of the projectile. Finally, the peak height of the projectile can be found using a time value which one-half the total time of flight. The most useful equation for this is usually y = viy*t +0.5*ay*t2 .

21. SUMMARY: See Hand-outs