2. 9.1.1State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field. 9.1.2Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance. 9.1.3Describe qualitatively the effect of air resistance on the trajectory of a projectile. 9.1.4Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion

3. 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). Topic 9: Motion in fields 9.1 Projectile motion FYI  We will ignore air resistance in the discussion that follows…

4. 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. Topic 9: Motion in fields 9.1 Projectile motion Slowing down in +y dir. Speeding up in -y dir. Constant speed in +x dir. a x = 0 a y = -g

5. 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! Topic 9: Motion in fields 9.1 Projectile motion

6. 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. Topic 9: Motion in fields 9.1 Projectile motion SKETCH POINTS  Peak to left of original one.  Pre-peak distance more than post-peak.

7. 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: Topic 9: Motion in fields 9.1 Projectile motion kinematic equations s = ut + (1/2)at 2 v = u + at a is constant Displacement Velocity kinematic equations ∆x = u x t + (1/2)a x t 2 v x = u x + a x t a x and a y are constant ∆y = u y t + (1/2)a y t 2 v y = u y + a y t

8. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion kinematic equations ∆x = u x t + (1/2)a x t 2 v x = u x + a x t a x and a y are constant ∆y = u y t + (1/2)a y t 2 v y = u y + a y t PRACTICE: Show that the reduced equations for projectile motion are SOLUTION:  a x = 0 in the absence of air resistance.  a y = -10 in the absence of air resistance. 0 0 reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t

9. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 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 = u x t becomes t = ∆x/u x so that t 2 = ∆x 2 /u x 2. Then ∆y = u y t - 5t 2, or ∆y = (u y /u x )∆x – (5/u x 2 )∆x 2. FYI  The equation of a parabola is y = Ax + Bx 2.  In this case, A = u y /u x and B = -5/u x 2. reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t

10. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t PRACTICE: A cannon fires a projectile with a muzzle velocity of 56 ms -1 at an angle of inclination of 15º. (a) What are u x and u y ? SOLUTION: Make a velocity triangle. u = 56 m s -1 u x = u cos  u y = u sin   = 15º u x = 56 cos 15º u x = 54 m s -1 u y = 56 sin 15º u y = 15 m s -1.

11. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t 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 u x = 54 and u y = 15: (c) At the maximum height, v y = 0. Why? Thus v y = 15 - 10t becomes 0 = 15 - 10t so that 10t = 15 t = 1.5 s. tailored equations for this particular projectile ∆x = 54t v x = 54 ∆y = 15t - 5t 2 v y = 15 - 10t

12. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t 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 t up = t down = 1.5 s so t = 3.0 s. Thus ∆x = 54t ∆x = 54(3.0) ∆x = 160 m.

13. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion reduced equations of projectile motion ∆x = u x t v x = u x ∆y = u y t - 5t 2 v y = u y - 10t 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, v x vs. t, and v y vs. t: SOLUTION:  The only acceleration is g in the – y-direction.  v x = 54, a constant. Thus it does not change over time.  v y = 15 - 10t Thus it is linear with a negative gradient and it crosses the time axis at 1.5 s. t ayay -10 t vxvx 54 t vyvy 15 1.5

14. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 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. v y = 0 at switch.

15. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  The flight time is limited by the y motion.  The maximum height is limited by the y motion.

16. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion a x = 0. a y = -10 ms -2.

17. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  Fall time limited by y-equations: ∆y = u y t - 5t 2 -33 = 0t - 5t 2 -33 = -5t 2 (33/5) = t 2  t = 2.6 s.

18. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  Use x-equations and t = 2.6 s: ∆x = u x t ∆x = 18(2.6)  ∆x = 15 m.

19. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion v x = u x v y = u y – 10t v x = 18. v y = 0 – 10t v y = –10(2.6) = -26. 18 26   tan  = 26/18  = tan -1 (26/18) = 55º.

20. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  The horizontal component of velocity is v x = u x which is CONSTANT.  The vertical component of velocity is v y = u y – 10t which is INCREASING (negatively).

21. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  ∆E K + ∆E P = 0 ∆E K = -∆E P ∆E K = -mg∆y ∆E K = -(0.44)(9.8)(-32) = + 138 J = E K – E K0 E K = + 138 + (1/2)(0.44)(22 2 ) = 240 J. E K0 = (1/2)mu 2

22. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  If 34% of the energy is consumed, 76% remains. 0.76(240) = 180 J (1/2)(0.44)v 2 = 180 J  v = 29 ms -1.

23. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  Use ∆E K + ∆E P = 0. (1/2)mv f 2 - (1/2)mv 2 = -∆E P mv f 2 = mv 2 + -2mg(0-H) v f 2 = v 2 + 2gH

24. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion u x = u cos  u x = 28 cos 30º u x = 24 m s -1. u y = u sin  u y = 28 sin 30º u y = 14 m s -1.

25. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion ∆x = u x t 16 = 24t t = 16/24 = 0.67 ∆y = u y t – 5t 2 ∆y = 14t – 5t 2 ∆y = 14(0.67) – 5(0.67) 2 = 7.1 m.  The time to the wall is found from ∆x…

26. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m u x = ∆x/∆t = (4-0)/(0.5-0.0) = 8 ms -1.

27. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m 11 m u y = ∆y/∆t = (11-0)/(0.5-0.0) = 22 ms -1.

28. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion 0.0s 0.5s 4 m 11 m 1.0s 1.5s 2.0s 2.5s 3.0s 24 m 30 m D 2 = 24 2 + 30 2 so that D = 38 m D   = tan -1 (30/24) = 51º,@  = 51º.

29. Solve problems on projectile motion. Topic 9: Motion in fields 9.1 Projectile motion  New peak below and left.  Pre-peak greater than post- peak.