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1 Projectile motion an object dropped from rest an object which is thrown vertically upwards an object is which thrown upwards at an angle A projectile.

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Presentation on theme: "1 Projectile motion an object dropped from rest an object which is thrown vertically upwards an object is which thrown upwards at an angle A projectile."— Presentation transcript:

1 1 Projectile motion an object dropped from rest an object which is thrown vertically upwards an object is which thrown upwards at an angle A projectile is an object upon which the only force acting is gravity. It is assumed that the influence of air resistance is negligible.

2 2 Monkey and Hunter Experiment

3 3 Only one force (weight) is acting on the cannon ball. Horizontal motion: Constant velocity WW W W W W vertical motion: Constant downward acceleration g

4 4 Horizontal motion: No horizontal force => constant velocity Vertical motion: Downward force (weight) => uniform downward acceleration (g)

5 5 A projectile motion  u Speed of projection: u Angle of projection:  Initial horizontal speed: u cos  Initial vertical speed: u sin  u cos  u sin  Horizontal displacement: x = u cos  t Vertical displacement y: = u sin  t – ½ gt 2

6 6 Time of flight Put y = 0 0 = u sin  t – ½ gt 2 t(u sin  – ½ gt) = 0 t = 0 (rejected) or t = 2u sin  / g t = 0 t = time of flight

7 7 Range Range = final horizontal displacement = horizontal speed x time of flight = u cos  x 2u sin  / g = 2 u 2 sin  cos  / g = u 2 sin 2  /g range

8 8 Maximum Range range Range = 2 u 2 sin  cos  / g = u 2 sin 2  /g Maximum range = u 2 /g when sin 2  = 1 i.e. 2  = 90 o  = 45 o

9 9 Summary Vertical motion: y = ut sin  – ½ gt 2 Horizontal motion: x = ut cos  Time of flight: Put y = 0 ⇒ t = 2usin  / g Range: Put t = 2u sin  / g ⇒ x = u 2 sin 2  / g

10 10 Exercise (HKAL 1998) page 28 A small object is thrown horizontally towards a vertical wall 1.2 m away. It hits the wall 0.8 m below its initial horizontal level. At what speed does the object hit the wall? (Neglect air resistance.) 1.2 m 0.8 m wall

11 11 Solution x = ut cos  Horizontal motion: x = ut cos  ⇒ 1.2 = ut cos 0 o ⇒ ut = 1.2 y = ut sin  – ½ gt 2 Vertical motion: y = ut sin  – ½ gt 2 ⇒ 0.8 = ut sin 0 o + ½ gt 2 ⇒ 0.8 = ½ (10)t 2 ⇒ t = 0.4 s 1.2 m 0.8 m wall u ∵ ut = 1.2 ∴ u = 1.2/0.4 = 3 ms -1 v = u + at At the wall, (v = u + at) Horizontal velocity = 3 + (0)(0.4) = 3 ms -1 Vertical velocity = 0 + (10)(0.4) = 4 ms -1 The speed at the wall = ( ) ½ = 5 ms -1

12 12 Class work page 13 (a)Time of flight: By s = ut + ½ at = 10 sin 30 o t + ½ (-10) t 2 5t 2 – 5t – 100 = 0 (t – 5)(t + 4) = 0 t = 5 s or t = -4 s (rejected) 10 ms o 100 m

13 13 Class work page 13 (b)Horizontal distance moved = u cos 30 o x time of flight = 10 cos 30 o x 5 = 43.3 m 10 ms o 100 m

14 14 Class work page 13 (c)Direction of flight: v x = 10 cos 30 o = 8.66 ms -1 v y = 10 sin 30 o + (-10)(5) = -45 ms -1 tan  = |v y / v x | = 45/8.66  = 79.1 o ∴ The particle makes an angle of 79.1 o to the horizontal when it reaches the sea. 10 ms o 100 m  v vyvy vxvx

15 15 Class work page 13 (d)Speed of the ball = sqrt(v x 2 + v y 2 ) = sqrt( ) = ms ms o 100 m  v vyvy vxvx

16 16 Homework Textbook page 51 (6, 7) Reference book page 116 (8, 12, 16) Due date: 2nd October Test: Chapter 4 projectile motion 2nd October


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