ProjectilesA “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 vH = d / t)

•Vertically: Acceleration at 9.8 ms-2 (use u v a s t)

Vertical

u =

v =

a =

s =

t =

Horizontal

vH =

d =

t =

Example 1A 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 flightb)The velocity of projectionc)The horizontal component of velocity as the ball hits the ground.d)The vertical component of velocity as the ball hits the grounde)The velocity of the ball as it hits the ground.

Example 1A 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 …

1st Sign convention and lists!

DOWN = +Vertical

u =

v =

a =

s =

t =

Horizontal

v =

d =

t =

?

3 m

0 ms-1

?

9.8 ms-2

1.25 m

?

?

Example 1A 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 at2

1.25 = 0 + (0.5x9.8) t2

1.25 = 4.9 t2

0.25 = t2

t = 0.5 s

Vertical

u =

v =

a =

s =

t =

0 ms-1

?

9.8 ms-2

1.25 m

?

Example 1A 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 = 3 0.5 = 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

vv= u + at = 0 + (9.8 x 0.5) = 4.9 ms-1

Vertical

u =

v =

a =

s =

t =

0 ms-1

?

9.8 ms-2

1.25 m

0.5s

e) The velocity of the ball as it hits the ground.

e)

Magnitude

x2 = vH2 + vv

2

= 62 + 4.92

x = 7.8 ms-1

6 ms-1

4.9 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 39o below the horizontal.

Often we must split the initial velocity into horizontal and vertical components, vH and uV.

Example 2

A ball is projected at an angle of 30o 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

vH = adj = 48 cos30o uV= opp = 48 sin30o

= 41.6 ms-1 = 24 ms-1

vH

uV48 ms-

1

30o

Vertical

u =

v =

a =

s =

t =

Horizontal

vH = 41.6 ms-1

d =

t =

lists

+ UP

?

24 ms-1

0 m/s at max height!

- 9.8 ms-2

?

?

?

c) v2 = u2 + 2as0 = 242 + (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

vH = 41.6 ms-1

d =

t =

lists

+ UP

?

24 ms-1

-24 ms-1 for full flight

- 9.8 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= vH x t= 41.6 x 4.9

= 203.8 m

Printed worked examples.

Revision questionsPage 12, questions 1 – 12.

Over to you – Problems 43 – 49.

Revision questionsPage 12, questions 1 – 12.

Problems 43 – 50.

43. a) 7.82 s b) 2346 m c) directly below plane44. a) 5 s b) 122.5 m45. b) 24.68 ms-1 at 53o from the horizontal

or 24.68 ms-1 at 37.4o from the vertical

46. a) (i) constant velocity (ii) constant downwards

acceleration b) 50 ms-1 at 370 from horizontal (or 53o 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) 141.2 m

48. a) 8 s b) 379 m49. a) 15.6 ms-1 b) assume no air resistance50. Describe how to measure acceleration.