2. Projectile Motion

3. A projectile is any object in free fall near the surface of the Earth

4. A projectile moves horizontally as well as vertically Examples: –A baseball, football, basketball, or soccer ball in flight –A bullet fired from a gun –A marble that has rolled off the table –A human long-jumper

5. The path of a projectile is a parabola

6. Components simplify projectile motion: 1)Displacements, velocities, and accelerations are resolved into x- and y-components x y V  x y VyVy VxVx

7. 2)The 1D-motion equations are applied to x- and y-directions independently No. Equation 1  x = ½(v x0 + v x )t 2v x = v x0 + a x t 3  x = v x0 t + ½ a x t 2 4 v x 2 = v x0 2 + 2a x  x No. Equation 1  y = ½(v y0 + v y )t 2v y = v y0 + a y t 3  y = v y0 t + ½ a y t 2 4 v y 2 = v y0 2 + 2a y  y x-direction: y-direction:

8. 3)The x- and y-components are recombined to determine the resultant motion x y VyVy VxVx x y V 

9. X-Direction in Projectile Motion We assume air resistance is zero Horizontal motion is constant velocity v x = v x0 = constant  x = v x t Constant velocity means acceleration is zero a x = 0 X1 X2 X3

10. Y-Direction in Projectile Motion We assume air resistance is zero Vertical motion is constant acceleration, a y = -g = -9.80 m/s 2  y = ½(v y0 + v y )t v y = v y0 - gt  y = v y0 t - ½ gt 2 v y 2 = v y0 2 - 2g  y Y1 Y2 Y3 Y4

11. Projectile Launched Horizontally Given: v x = constant = v x0 v y0 = 0  y = -h Unknowns: t = ?  x = ? v = ? vxvx h v

12. How long will it be airborne? Use Eqn. Y3: -h = -½ gt 2 t = (2h/g) ½ How far will it go? Use Eqn. X2:  x = v x t How fast will it hit? Use Eqn. Y4: v y 2 = 2gh v 2 = v x 2 + v y 2 v 2 = v x 2 + 2gh H1 H2 H3

13. Example: Assume Wile E. Cayote fires the cannonball horizontally at 25.2 m/s at a height of 7.91 m: t = (2h/g) ½ = [2(7.91 m)/(9.80 m/s 2 )] ½ = 1.27 s  x = v x t = (25.2 m/s)(1.27 s) = 32.0 m v = (v x 2 + 2gh) ½ = [(25.2m/s) 2 + 2(9.80m/s 2 )(7.91m)] ½ = 28.1 m/s

14. Projectile Launched at an Angle Given: v x = v 0 cos  v y0 = v 0 sin   y = 0 Unknowns: t = ?  x = ? v = ? h = ? v0v0  h v xx v0v0 v y0 vxvx 

15. How long will it be airborne? Use Eqn. Y3: 0 = v 0 sin  ·t – (½)gt 2 t = (2v 0 /g)sin  How far will it go? Use Eqn. X2:  x = v x t = (v 0 cos  )t  x = (2v 0 2 /g)cos  sin  How fast will it hit? Use Eqn. Y1: 0 = ½(v y0 + v y )t v y = -v y0 v = v 0 (in magnitude) A1 A3 A2

16. How high will it go? Use Eqn. Y4 (half way): So  y = h and v y = 0 v y 2 = v y0 2 - 2g  y 0 = v y0 2 - 2gh h = v y0 2 /(2g) h = (v 0 sin  ) 2 /(2g) V0V0  h A4

17. Example: A golf ball is hit at an initial velocity of 43.9 m/s at an angle of 33.3  from horizontal: v x = v 0 cos  = (43.9 m/s)cos(33.4  ) = 36.6 m/s v y0 = v 0 sin  = (43.9 m/s)sin(33.4  ) = 24.2 m/s t = (2v 0 /g)sin  = [2(43.9 m/s)/(9.80 m/s 2 )]sin(33.4  ) = 4.93 s  x = v x t = (2v 0 2 /g)cos  sin  = [2(43.9 m/s) 2 /(9.80 m/s 2 )]cos(33.4  )sin(33.4  ) = 181 m h = v y0 2 /(2g) = (v 0 sin  ) 2 /(2g) = [(43.9 m/s)sin(33.4  )] 2 /(9.80 m/s 2 )/2 = 29.8 m