1 / 21

# Projectile Motion

Projectile Motion. Projectile Motion. Today’s Objectives : Recognize examples of projectile motion, Recognize that the horizontal and vertical components of a projectile’s motion are independent of each other, and

## Projectile Motion

E N D

### Presentation Transcript

1. Projectile Motion

2. Projectile Motion Today’s Objectives: • Recognize examples of projectile motion, • Recognize that the horizontal and vertical components of a projectile’s motion are independent of each other, and • Resolve vectors into their components, and apply the kinematic equations to solve problems involving projectiles.

3. What Is Projectile Motion? Projectile Motion is the motion of objects moving in two dimensions under the influence of gravity. Air resistance is negligible (we can ignore it).

4. EXAMPLES OF PROJECTILES Examples of projectiles include anything launched or thrown into the air such as cannon balls, baseballs, pumpkins, soccer balls, tennis balls, handballs, racquetballs, people, four-square balls, softballs, mothballs, fireworks, water from rain clouds or from a hose…

5. There Are Some Common Misconceptions…. Going fast horizontally means you don’t fall as fast. If you go fast enough, you don’t fall at all. Gravity won’t act on you until you look down.

6. The Truth Is... If gravity is the only force acting on an object, it will accelerate at a rate of 9.8 m/s2 DOWN, regardless of what’s happening horizontally. In fact, if the object doesn’t have wings, jet engines, propellers or rockets, its horizontal motion will have absolutely no effect on its vertical motion. They are completely independent of each other.

7. This is Key… …To understanding projectiles: Horizontally, projectiles move with constant velocity. Vertically, projectiles move with constant acceleration.

8. WHY? Lets Consider a Little Information About… Newton’s Laws of Motion Newton’s First Law of Motion, also known as the Law of Inertia, states that a body in motion will remain in motion at constant velocity, unless acted upon by an unbalanced force (and that a body at rest will remain at rest). Therefore, if there were no gravity, an object fired from a cannon off a hilltop would continue along a straight-line path at constant velocity forever and ever…the horizontal motion remains constant.

9. The Path of a Projectile… Without Gravity The cannon ball moves a constant amount with each second.

10. WATCH THE NUMBERS

11. The Horizontal Motion is Independent of the Vertical Motion

12. HORIZONTAL VELOCITY is STILL Constant

13. What’s a Satellite??

14. Vertical Components: y - vertical displacement g - vertical acceleration (9.8 m/s2) vyi - initial vertical velocity vyf- final vertical velocity t - time Horizontal Components:  x - horizontal displacement vxi = vxf-horizontal velocity t - time Consider Horizontal and Vertical Components of Motion Independently How Do We Solve Projectile Problems??

15. Vertical Equations vf =vi + at vyf =vyi - gt vf2 =vi2 + 2a y vyf2 =vyi2 - 2g y y = vit+ 1/2at2 y = vyit– 1/2gt2 y = 1/2(vi+vf)t y = 1/2(vyi+vyf)t y = vft- 1/2at2 y = vit+ 1/2gt2 Horizontal Equation x = vxt How Do We Solve Projectile Problems?? These equations assume that down is negative!! Since horizontal acceleration is zero, this is the only equation.

16. How Do We Solve Projectile Problems?? • Draw an accurate diagram showing the trajectory of the object. • Solve for the horizontal and vertical components of the initial velocity, if given. • Complete a data table using given and implied data, based on an appropriate sign convention (you assign). • Identify what you are looking for. Remember you need at least 3 pieces of vertical data to use kinematic equations… if you don’t have 3, look to the horizontal data to find time, which is the only common variable.

17. ‘Human’ Cannonball Mr. Maroo is launched from a cannon with an initial velocity of 18 m/s at an angle of 25° with the horizontal. (a) How high will he go? (b) Where should a safety net be placed so that he lands safely? (c) How much time does he spend in the air? V H Given: Vi=18 m/s Vxi=16.3 m/s Vyi=7.6 m/s g=-9.8 m/s2 Vyi=18 m/s sin 25°= 7.6 m/s 25° Vyf= 0 m/s Vxi=18 m/s cos 25°= 16.3 m/s

18. Practice With an Applet http://galileo.phys.virginia.edu/classes/109N/more_stuff/Applets/ProjectileMotion/jarapplet.html

More Related