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Topic 9: Motion in fields 9.1 Projectile motion

Topic 9: Motion in fields 9.1 Projectile motion. 9.1.1 State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

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Topic 9: Motion in fields 9.1 Projectile motion

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  1. Topic 9: Motion in fields9.1 Projectile motion 9.1.1 State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2 Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. 9.1.3 Describe qualitatively the effect of air resistance on the trajectory of a projectile. 9.1.4 Solve problems on projectile motion.

  2. Topic 9: Motion in fields9.1 Projectile motion State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity. Baseballs, stones, or bullets are all examples of projectiles. You know that all objects moving through air feel an air resistance (recall sticking your hand out of the window of a moving car). FYI We will ignore air resistance in the discussion that follows…

  3. Topic 9: Motion in fields9.1 Projectile motion State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. Regardless of the air resistance, the vertical and the horizontal components of velocity of an object in projectile motion are independent. Speeding up in -y dir. Slowing down in +y dir. ay= -g ay= -g ax= 0 Constant speed in +x dir.

  4. Topic 9: Motion in fields9.1 Projectile motion Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. The trajectory of a projectile in the absence of air is parabolic. Know this!

  5. Topic 9: Motion in fields9.1 Projectile motion Describe qualitatively the effect of air resistance on the trajectory of a projectile. If there is air resistance, it is proportional to the square of the velocity. Thus, when the ball moves fast its deceleration is greater than when it moves slow. SKETCH POINTS Peak to left of original one. Pre-peak distance more than post-peak.

  6. kinematic equations s = ut + (1/2)at2 Displacement Velocity v = u + at a is constant kinematic equations ∆x = uxt + (1/2)axt2 vx = ux + axt ax and ayare constant ∆y = uyt + (1/2)ayt2 vy = uy + ayt Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. Recall the kinematic equations from Topic 2: Since we worked only in 1D at the time, we didn’t have to distinguish between x and y in these equations. Now we appropriately modify the above to meet our new requirements of simultaneous equations:

  7. 0 0 reduced equations of projectile motion ∆x = uxt ∆y = uyt - 5t2 vx = ux vy = uy - 10t kinematic equations ∆x = uxt + (1/2)axt2 vx = ux + axt ax and ayare constant ∆y = uyt + (1/2)ayt2 vy = uy + ayt Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. PRACTICE: Show that the reduced equations for projectile motion are SOLUTION: ax = 0 in the absence of air resistance. ay = -10 in the absence of air resistance.

  8. reduced equations of projectile motion ∆x = uxt ∆y = uyt - 5t2 vx = ux vy = uy - 10t Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. EXAMPLE: Use the reduced equations above to prove that projectile motion (in the absence of air resistance) is parabolic. SOLUTION: Just solve for t in the first equation and substitute it into the second equation. ∆x = uxt becomes t = ∆x/uxso that t2 = ∆x2/ux2. Then ∆y = uyt - 5t2, or ∆y = (uy/ux)∆x – (5/ux2)∆x2. FYI The equation of a parabola is y = Ax + Bx2. In this case, A = uy/uxand B = -5/ux2.

  9. reduced equations of projectile motion ∆x = uxt ∆y = uyt - 5t2 vx = ux vy = uy - 10t Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (a) What are ux and uy? SOLUTION: Make a velocity triangle. u = 56 ms-1 uy= usin = 15º uy= 56sin15º ux= ucos uy= 15 ms-1. ux= 56cos15º ux= 54 ms-1

  10. reduced equations of projectile motion tailored equations for this particular projectile ∆x = uxt ∆x = 54t ∆y = uyt - 5t2 ∆y = 15t - 5t2 vx = ux vx = 54 vy = uy - 10t vy = 15 - 10t Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (b) What are the tailored equations of motion? (c) When will the ball reach its maximum height? SOLUTION: (b) Just substitute ux = 54 and uy = 15: (c) At the maximum height, vy = 0. Why? Thus vy = 15 - 10t becomes0 = 15 - 10t so that 10t = 15 t = 1.5 s.

  11. reduced equations of projectile motion ∆x = uxt ∆y = uyt - 5t2 vx = ux vy = uy - 10t Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (d) How far from the muzzle will the ball be when it reaches the height of the muzzle at the end of its trajectory? SOLUTION: From symmetry tup = tdown = 1.5 s so t = 3.0 s. Thus ∆x = 54t ∆x = 54(3.0) ∆x = 160 m.

  12. reduced equations of projectile motion ∆x = uxt ∆y = uyt - 5t2 vx = ux vy = uy - 10t vy ay vx t t t -10 54 15 1.5 Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 15º. (e) Sketch the following graphs: a vs. t, vx vs. t, and vy vs. t: SOLUTION: The only acceleration is g in the –y-direction. vx = 54, a constant. Thus it does not change over time. vy = 15 - 10t Thus it is linear with a negative gradient and it crosses the time axis at 1.5 s.

  13. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. The acceleration is ALWAYS g for projectile motion-since it is caused by Earth and its field. At the maximum height the projectile switches from upward to downward motion. vy = 0 at switch.

  14. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. The flight time is limited by the y motion. The maximum height is limited by the y motion.

  15. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. ax = 0. ay = -10 ms-2.

  16. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. ∆y = uyt - 5t2 -33 = 0t - 5t2 -33 = -5t2 (33/5) = t2 Fall time limited by y-equations: t = 2.6 s.

  17. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. ∆x = uxt ∆x = 18(2.6) Use x-equations and t = 2.6 s: ∆x = 15 m.

  18. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. 18  vx = ux vx = 18. 26 vy = uy– 10t vy = 0– 10t vy = –10(2.6) = -26. tan = 26/18  = tan-1(26/18) = 55º.

  19. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. The horizontal component of velocity is vx = uxwhich is CONSTANT. The vertical component of velocity is vy = uy – 10twhich is INCREASING (negatively).

  20. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion.  ∆EK + ∆EP= 0 ∆EK = -∆EP ∆EK = -mg∆y EK0 = (1/2)mu2 ∆EK = -(0.44)(9.8)(-32) = +138 J = EK – EK0 EK = +138 + (1/2)(0.44)(222) = 240 J.

  21. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. If 34% of the energy is consumed, 76% remains. 0.76(240) = 180 J (1/2)(0.44)v2 = 180 J v = 29 ms-1.

  22. (1/2)mvf2 - (1/2)mv2 = -∆EP mvf2 = mv2 + -2mg(0-H) vf2 = v2 + 2gH Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. Use ∆EK + ∆EP= 0.

  23. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. uy= usin uy= 28sin30º ux= ucos ux= 28cos30º ux= 24 ms-1. uy= 14 ms-1.

  24. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. ∆x = uxt 16 = 24t t = 16/24= 0.67 The time to the wall is found from ∆x… ∆y = uyt – 5t2 ∆y = 14t – 5t2 ∆y = 14(0.67) – 5(0.67)2 = 7.1 m.

  25. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. 0.5s 0.0s 4 m ux = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms-1.

  26. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. 0.5s 11 m 0.0s 4 m uy = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms-1.

  27. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. 2.5s 3.0s 2.0s 30 m 1.5s 1.0s D 0.5s 11 m  = tan-1(30/24) = 51º 0.0s  4 m 24 m D2 = 242 + 302 so that D = 38 m ,@  = 51º.

  28. Topic 9: Motion in fields9.1 Projectile motion Solve problems on projectile motion. New peak below and left. Pre-peak greater than post-peak.

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