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PROJECTILE MOTION

PROJECTILE MOTION. Qiu Cheng Catherine Brostrom Michelle Wood Honors Physics. Therefore you must have two lists: Y list X list. Horizontally Launched Projectiles. Non-Horizontally Launched Projectiles.

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PROJECTILE MOTION

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  1. PROJECTILE MOTION Qiu Cheng Catherine Brostrom Michelle Wood Honors Physics

  2. Therefore you must have two lists: Y list X list

  3. Horizontally Launched Projectiles

  4. Non-Horizontally Launched Projectiles

  5. Suppose a rescue airplane drops a relief package while it is moving with a constant horizontal speed at an elevated height. Assuming that air resistance is negligible, where will the relief package land relative to the plane? a. below the plane and behind it. b. directly below the plane c. below the plane and ahead of it

  6. 1. Consider these diagrams in answering the following questions. Which diagram (if any) might represent ... a. ... the initial horizontal velocity? b. ... the initial vertical velocity? c. ... the horizontal acceleration? d. ... the vertical acceleration? e. ... the net force? The initial horizontal velocity is A (It's the only horizontal vector). The initial vertical velocity could be B (if projected down) or C (if projected upward). None of these; there is no horizontal acceleration. The vertical acceleration is B; it is always downwards. The net force on a projectile is B (there is only one force - gravity; and it is downwards).

  7. Solving Projectile Motion Problems

  8. Projectile Motion Equations D y = vit + ½(a)(t)2 vf = vi + (a)(t) vf2 = vi2 + 2(a)(D)

  9. Trig Functions Sinq = opp./hyp. Cosq = adj./hyp. Tanq = opp./adj.

  10. Type 1 Problems A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

  11. What You Know... Now Pick your Equations... y = viy • t + 0.5 • ay • t2 & x = vix • t + 0.5 • ax • t2

  12. Solution • y = vit + ½(a)(t)^2 -22.0=0(t) + ½ (-9.8)(t)^2 t= 2.12sec • x = vit + ½(a)(t)^2 35.0= vix(2.12)+ ½ (0)^2 16.5 m/s.

  13. Try This One On Your Own... A potato cannon fires a potato horizontally from the edge of the roof of a 7.347m tall building. The potato lands 19.592m from the building in a lot that is level with the base of the building. What is the launch velocity of the potato? 7.347m 19.592m

  14. Lists X DX= 19.592 a=0 vi=? t= Y DY= -7.347 a=-9.8 vi=0 t=

  15. Finding Time Vf^2=Vi^2+2a(Dy) Vf^2=0^2+2(-9.8)(-7.347) Vf^2=144.0012 Vf= +/- 12.00 Vf =Vi+at 12=0+(-9.8)(t) t=-1.22 Vf =Vi+at -12=0+(-9.8)(t) t=1.22

  16. Find Launch Velocity DX= Vit+ ½ a(t)^2 19.592=Vi(1.22) + ½ 0(1.22)^2 19.592=1.22Vi Vi=16.06m/s

  17. Finding the Initial Velocity Components Consider a projectile launched with an initial velocity of 50 m/s at an angle of 60 degrees above the horizontal……

  18. Examples 1) A water balloon is launched with a speed of 40 m/s at an angle of 60 degrees to the horizontal. 2) A motorcycle stunt person traveling 70 mi/hr jumps off a ramp at an angle of 35 degrees to the horizontal. 3) A springboard diver jumps with a velocity of 10 m/s at an angle of 80 degrees to the horizontal.

  19. Solutions 1) 2) 3) 70mi/hr 10m/s 40m/s 60 35 80 cos (60 deg) = vx / (40 m/s) vx = 40 m/s • cos (60 deg) = 20.0 m/s cos (80 deg) = vx / (10 m/s) vx = 10 m/s • cos (80 deg) = 1.7 m/s cos (35 deg) = vx / (70 mi/hr) vx = 70 mi/hr • cos (35 deg) = 57.3 mi/hr sin (80 deg) = vy / (10 m/s) vy = 10 m/s • sin (80 deg) = 9.8 m/s sin (60 deg) = vy / (40 m/s) vy = 40 m/s • sin (60 deg) = 34.6 m/s sin (35 deg) = vy / (70 mi/hr) vy = 70 mi/hr • sin (35 deg) = 40.1 mi/hr

  20. Type 2 Problems After eating raw cookie dough, a student catches salmonella. When standing, the student’s mouth is 1.5m off the floor. If she tilts her heard back, and expels the contents of her stomach at 5m/s, 36.9° above horizontal, how far from her feet (what is the range) does the partially digested food land? 36.9° 5m/s

  21. Velocity Componets Vix- cos36.9°=Vix/5m/s Vix=3.99 Viy- sin36.9°=Viy/5m/s Viy=3.00 5m/s Viy 36.9° Vix

  22. Lists X DX=? a=0 vi=3.99 t=? Y DY= -1.5 a= -9.8 vi= 3.00 t=?

  23. Pick Your Equation D y = vit + ½(a)(t)^2 vf = vi + (a)(t) vf2 = vi2 + 2(a)(D)

  24. Plug In Vf^2 = vi^2 + 2a(Dy) Vf^2= (3.00)^2+2(-9.8)(-1.5) Vf^2=38.6 Vf = +/- 6.2

  25. Pick Your Equation D y = vit + ½(a)(t)^2 vf = vi + (a)(t) vf2 = vi2 + 2(a)(D)

  26. Plug In vf = vi + (a)(t) -6.2= 3.00+ (-9.8)(t) t= .938 6.2= 3.00+ (-9.8)(t) t=-.327

  27. Pick Your Equation D x = vit + ½(a)(t)^2 vf = vi + (a)(t) vf2 = vi2 + 2(a)(D)

  28. Plug In D x = vit + ½(a)(t)^2 DX= 3.99(.938) + ½(0)(.938) DX= 3.74262

  29. Try This One... 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. 25m/s 45°

  30. Velocity Componets vix = 25 m/s•cos(45 deg) vix = 17.7 m/s 25m/s Vif viy = 25 m/s•sin(45 deg) viy = 17.7 m/s 45° Vix

  31. Lists X DX= ? Vi= 17.77 Vf= 17.77 a=0 Y DY=? Vi=17.77 Vf=-17.77 a=-9.8 *As indicated in the table, the final x-velocity (vfx) is the same as the initial x-velocity (vix). This is due to the fact that the horizontal velocity of a projectile is constant; there is no horizontal acceleration. The table also indicates that the final y-velocity (vfy) has the same magnitude and the opposite direction as the initial y-velocity (viy). This is due to the symmetrical nature of a projectile's trajectory.

  32. Equation vfy = viy + ay*t -17.7 m/s = 17.7 m/s + (-9.8 m/s/s)•t 3.61 s = t

  33. Equation x = vix•t + 0.5•ax•t2 x = (17.7 m/s)•(3.6077 s) + 0.5•(0 m/s/s)•(3.6077 s)2 x = 63.8 m

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