Projectiles A “projectile” is an object which is projected (thrown, dropped, fired) into the air or space. e.g. marble pushed off the edge of a bench. Constant Speed Constant acceleration of 9.8 ms -2 e.g. Cannon ball being fired into the air. Max Height At max height, vertical v = 0 ms -1 Upwards journey is the exact opposite of downward journey. These Objects travel in a curved path! NOTES p.16

A projectile moves horizontally and vertically simultaneously so, for calculations, we need to separate these motions. Horizontally: Constant velocity (use v H = d / t) Vertically: Acceleration at 9.8 ms -2 (use u v a s t) Vertical u = v = a = s = t = Horizontal v H = d = t =

Example 1 A ball is projected horizontally off the end of a bench. It hits the floor 3 m from the base of the bench which is 1.25m high. Find a)The time of flight b)The velocity of projection c)The horizontal component of velocity as the ball hits the ground. d)The vertical component of velocity as the ball hits the ground e)The velocity of the ball as it hits the ground.

Example 1 A ball is projected horizontally off the end of a bench. It hits the floor 3 m from the base of the bench which is 1.25 m high. Find … 1 st Sign convention and lists! DOWN = + Vertical u = v = a = s = t = Horizontal v = d = t = ? 3 m 0 ms -1 ? 9.8 ms m ? ?

Example 1 A ball is projected horizontally off the end of a bench. It hits the floor 3 m from the base of the bench which is 1.25m high. Find a)The time of flight a) Vertical s = ut + ½ at = 0 + (0.5x9.8) t = 4.9 t = t 2 t = 0.5 s Vertical u = v = a = s = t = 0 ms -1 ? 9.8 ms m ?

Example 1 A ball is projected horizontally off the end of a bench. It hits the floor 3 m from the base of the bench which is 1.25m high. Find b)The velocity of projection c)The horizontal component of velocity as the ball hits the ground. b) Horizontal v = d t = = 6 ms -1 Horizontal v = d = t = ? 3 m 0.5s c) Horizontal component of velocity on landing = 6 ms -1 as this remains constant throughout the flight.

d)The vertical component of velocity as the ball hits the ground d) Vertical v v = u + at = 0 + (9.8 x 0.5) = 4.9 ms -1 Vertical u = v = a = s = t = 0 ms -1 ? 9.8 ms m 0.5s

e)The velocity of the ball as it hits the ground. e) Magnitude x 2 = v H 2 + v v 2 = x = 7.8 ms -1  6 ms ms -1 x Direction tan  opp adj = 4.9/6  = 39 o This is the resultant of the horizontal & vertical components! So landing velocity = 7.8 ms -1 at 39 o below the horizontal.

Often we must split the initial velocity into horizontal and vertical components, v H and u V. Example 2 A ball is projected at an angle of 30 o the horizontal, with a velocity of 48 ms -1. Calculate a)the horizontal component of the initial velocity. b)the vertical component of this initial velocity. c)the maximum height d)the time of flight. e)the range. a) horizontal b) vertical v H = adj = 48 cos30 o u V = opp = 48 sin30 o = 41.6 ms -1 = 24 ms -1 vHvH uVuV 48 ms o

Vertical u = v = a = s = t = Horizontal v H = 41.6 ms -1 d = t = lists + UP ? 24 ms -1 0 m/s at max height! ms -2 ? ? ? c) v 2 = u 2 + 2as 0 = (2 x -9.8 x s) 0= 576 – 19.6 s 19.6 s = 576 s = 29.4 m s = 576 / 19.6

Vertical u = v = a = s = t = Horizontal v H = 41.6 ms -1 d = t = lists + UP ? 24 ms ms -1 for full flight ms -2 0 m for full flight ? ? d) v = u + at -24 = 24 + (-9.8 t) -48 = -9.8t t= 48/9.8 = 4.9s e) d= v H x t = 41.6 x 4.9 = m

Printed worked examples. Revision questions Page 12, questions 1 – 12. Over to you – Problems 43 – 49.

Revision questions Page 12, questions 1 – 12. Problems 43 – a) 7.82 s b) 2346 m c) directly below plane 44. a) 5 s b) m 45. b) ms -1 at 53 o from the horizontal or ms -1 at 37.4 o from the vertical 46. a) (i) constant velocity (ii) constant downwards acceleration b) 50 ms -1 at 37 0 from horizontal (or 53 o from vertical) c) 40 ms -1 d) 45 m e) 240 m 47. a) 20 ms -1 b) 20.4 m c) 4.08 s d) m 48. a) 8 s b) 379 m 49. a) 15.6 ms -1 b) assume no air resistance 50. Describe how to measure acceleration.