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Prior knowledge check1) A small ball is projected vertically

upwards from a point P with speed 15ms-1. The ball is modelled as a particle moving freely under gravity. Find:

a) The maximum height of the ball

b) The time taken for the ball to return to P

2) Write expressions for x and y in terms of v and θ, based on the diagram below

3a) Given that 𝑠𝑖𝑛𝜃 =5

13and 𝜃 is acute,

find 𝑐𝑜𝑠𝜃 and 𝑡𝑎𝑛𝜃.

b) Given that 𝑡𝑎𝑛𝜃 =8

15and 𝜃 is reflex,

find 𝑠𝑖𝑛𝜃 and 𝑐𝑜𝑠𝜃.

11.5𝑚

3.1𝑠

𝑥 = 𝑣𝑐𝑜𝑠𝜃

v

x

y

θ

𝑦 = 𝑣𝑠𝑖𝑛𝜃

𝑐𝑜𝑠𝜃 =12

13𝑡𝑎𝑛𝜃 =

5

12

𝑠𝑖𝑛𝜃 = −8

17𝑐𝑜𝑠𝜃 = −

15

17

You can use the constant acceleration formulae for a projectile moving in a

vertical plane

A ball is thrown horizontally, with speed 20ms-1, from the top of a building of

height 30m.

Find:

a) The time the ball takes to reach the ground

b) The horizontal distance travelled in that time

20ms-1

30

m

As the ball is projected horizontally, there is no initial velocity vertically

Resolving vertically (with downwards as the positive direction) The ball must travel 30m to hit the ground

𝑠 = 30 𝑢 = 0 𝑣 =? 𝑎 = 9.8 𝑡 =?

30 = (0)(𝑡) +1

2𝑎(9.8)𝑡2

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

30 = 4.9𝑡2

2.5 = 𝑡 (2sf)

Sub in values

Work out terms

Calculate the positive value

𝑡 = 2.5

Start with a diagram!

Projectiles

6A

(𝑅 ↓)


vertical plane

A ball is thrown horizontally, with speed 20ms-1, from the top of a building of

height 30m.

Find:

a) The time the ball takes to reach the ground

b) The horizontal distance travelled in that time

The horizontal speed will remain constant at 20ms-1

30

m


𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 × 𝑡𝑖𝑚𝑒

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 20 × 2.5

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 49𝑚 (2sf)

Make sure you use the exact value for t rather than the rounded one from part a)!

𝑠 = 49𝑚

𝑡 = 2.5

Projectiles

6A

(𝑅 →)

20ms-1


vertical plane

A particle is projected horizontally with a velocity of 15ms-1. Find:

a) The horizontal and vertical components of the displacement of

the particle from the point of projection after 3 seconds

So we need to find how far the object has moved horizontally after 3

seconds, and how far it has moved vertically

The horizontal speed is constant…

Projectiles

6A

15ms-1Start with a diagram!

You can use 𝑦 and 𝑥 to represent the vertical and horizontal displacements…

𝑦

𝑥

𝑡 = 3𝑠

𝑢𝑥 = 15 𝑢𝑦 = 0

(𝑅 →)

𝑑 = 𝑠𝑡

𝑥 = (15)(3)

𝑥 = 45𝑚

Sub in values – remember that 𝑥 is the horizontal

distance

Calculate

𝑥 = 45𝑚

The initial velocities in the horizontal and vertical directions are sometimes represented by 𝑢𝑥 and 𝑢𝑦 respectively…


vertical plane




So we need to find how far the object has moved horizontally after 3

seconds, and how far it has moved vertically

We need to use a suvat equation to find the vertical distance, since the

velocity is not constant…

Projectiles

6A



𝑦

𝑥

𝑡 = 3𝑠

𝑢𝑥 = 15 𝑢𝑦 = 0

(𝑅 ↓)

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

Sub in values

𝑥 = 45𝑚

The initial velocities in the horizontal and vertical directions are sometimes represented by 𝑢𝑥 and 𝑢𝑦 respectively…

𝑦 = (0)(3) +1

2(9.8)(3)2

𝑦 = 44.1𝑚

Calculate

𝑠 = 𝑦 𝑢 = 0 𝑣 =? 𝑎 = 9.8 𝑡 = 3

𝑦 = 44.1𝑚


vertical plane




b) Find the distance of the particle from its starting point after 3 seconds

We know that the particle has moved 45m horizontally, and 44.1m vertically

downwards…

The total distance can be found using Pythagoras’ Theorem…

Projectiles

6A



𝑦

𝑥

𝑡 = 3𝑠

𝑥 = 45𝑚

𝑦 = 44.1𝑚

45𝑚

44.1𝑚𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒

𝑑 = (45)2+(44.1)2

= 63𝑚 (2sf)


vertical plane

A particle is projected horizontally with a speed of 𝑈 𝑚𝑠−1 from a point 122.5m above a horizontal plane. The particle hits the plane at a point which is at a horizontal distance of 90m away from

the starting point.

Find the initial speed of the particle.

Remember that the horizontal velocity is constant…

Therefore:

Projectiles

6A

𝑈 ms-1

Start with a diagram and label all the information you have…

90𝑚

122.5𝑚

𝑑 = 𝑠𝑡

90 = 𝑢𝑡

We know that the distance travelled is 90m

So if we can find the time taken, we can then use it to find the initial velocity…

We can use the vertical information to find when the particle hits the plane…


vertical plane


the starting point.


Projectiles

6A

𝑈 ms-1


90𝑚

122.5𝑚

(𝑅 ↓)

𝑠 = 112.5 𝑢 = 0 𝑣 =? 𝑎 = 9.8 𝑡 =?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

112.5 = (0)(𝑡) +1

2(9.8)𝑡2

112.5 = 4.9𝑡2

We can use the vertical information to find when the particle hits the plane…

5 = 𝑡

Sub in values

Simplify right side

Calculate, and take the positive value of 𝑡

𝒕 = 𝟓


vertical plane


the starting point.


Projectiles

6A

𝑈 ms-1


90𝑚

122.5𝑚

𝒕 = 𝟓

𝑑 = 𝑠𝑡

90 = 𝑢(5)

We know that the distance travelled is 90m, after 5 seconds. The speed is 𝑢.

(𝑅 →)

Calculate18 = 𝑢

Now we can resolve horizontally…

So the initial projection speed is 18ms-1

You need to understand how to resolve initial projection velocities that are

given at an angle (usually relative to a horizontal plane)

When a particle is projected with initial speed 𝑢, at an angle 𝜃 to the horizontal,

it moves along a symmetrical curve

The speed 𝑢 is known as the speed of projection

The angle 𝜃 is known as the angle of projection (or angle of elevation) of the

particle

The initial projection speed can be resolved into two components as shown

to the right

𝑢 𝑚𝑠−1

θ

𝑢 𝑚𝑠−1

θ

𝑢𝐶𝑜𝑠𝜃 𝑚𝑠−1

𝑢𝑆𝑖𝑛𝜃 𝑚𝑠−1

Any particle projected at an angle will have an initial horizontal and vertical speed, which you will usually

need to resolve separately…

Projectiles

6B



A particle is projected from a point on a horizontal plane with an initial velocity of 40ms-1, at an angle of 𝛼 above the

horizontal, where 𝑡𝑎𝑛𝛼 =3

4.

a) Find the horizontal and vertical components of the velocity

Start by finding the corresponding ratios for sine and cosine

b) Write the initial velocity in vector form

Projectiles

6B

3

4

5

𝑡𝑎𝑛𝛼 =3

4

𝑠𝑖𝑛𝛼 =3

5

𝑐𝑜𝑠𝛼 =4

5

𝛼

40 𝑚𝑠−1

𝛼

40𝑐𝑜𝑠𝛼 𝑚𝑠−1

40𝑠𝑖𝑛𝛼 𝑚𝑠−1

We can now use these ratio to split the initial velocity into its horizontal and vertical component parts

32 𝑚𝑠−1

24 𝑚𝑠−1

Calculate the exact values using the ratio we found…

32𝒊 + 24𝒋 𝑚𝑠−1



A particle is projected with velocity 𝑼 = 3𝒊 + 5𝒋 𝑚𝑠−1, where 𝒊 and 𝒋 are the

unit vectors in the horizontal and vertical directions respectively.

Find the initial speed of the particle and its angle of projection.

Projectiles

6B

𝜃

3𝒊

5𝒋

Initial speed – use Pythagoras’ Theorem…

= (3)2+(5)2

= 34

34

Angle of projection – use tan…

= 𝑇𝑎𝑛−15

3

= 59° (2sf)

So the initial projection is at speed 34 𝑚𝑠−1 at an angle of 59° above the horizontal

59°


vertical plane

A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30°. After projection, the particle moves freely

under gravity until it strikes the plane at a point A.

Find:

a) The greatest height above the plane reached by P

b) The time of flight of P

c) The distance OA


O A

28

30°

28cos30

28sin30

The greatest height will be reached when the vertical velocity is 0 (as the particle stops moving up and starts

moving down)

Resolve vertically and use SUVAT

𝑠 =? 𝑢 = 28𝑠𝑖𝑛30 𝑣 = 0 𝑎 = −9.8 𝑡 =?

𝑣2 = 𝑢2 + 2𝑎𝑠

(0)2= (28𝑠𝑖𝑛30)2+2(−9.8)(𝑠)

0 = 196 − 19.6𝑠

19.6𝑠 = 196

𝑠 = 10𝑚

Sub in values

Calculate terms

Rearrange

Calculate

10𝑚

(𝑅 ↑)

Projectiles

6C


vertical plane



Find:



c) The distance OA


O A

28

30°

28cos30

28sin30

The time of flight of the particle will be when its vertical displacement is 0 (ie – it is at the same height

it started at)

Resolve vertically and use SUVAT (again)

𝑠 = 0 𝑢 = 28𝑠𝑖𝑛30 𝑣 =? 𝑎 = −9.8 𝑡 =?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

0 = (28𝑠𝑖𝑛30)(𝑡) +1

2(−9.8)(𝑡2)

0 = 28𝑡𝑠𝑖𝑛30 − 4.9𝑡2

0 = 𝑡(28𝑠𝑖𝑛30 − 4.9𝑡)

𝑡 = 0 𝑜𝑟 28𝑠𝑖𝑛30 − 4.9𝑡 = 0

𝑡 = 2.9

Sub in values

Work out terms

Factorise

Either part could be 0

Work out the second possibility

The answer of 0 corresponds to the particle’s starting position. The value 2.9 is the time it

takes to hit the ground again

2.9𝑠

10𝑚

(𝑅 ↑)

Projectiles

6C


vertical plane



Find:



c) The distance OA


O A

28

30°

28cos30

28sin30

The distance OA will be based on the horizontal projection speed (which is constant)

We can use the time of flight calculated in part b)


𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 28𝑐𝑜𝑠30 × 2.9

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 69𝑚 (2𝑠𝑓)

69𝑚

2.9𝑠

10𝑚

Projectiles

6C

You can use the constant acceleration formulae for a projectile moving in a vertical

plane

A particle is projected from a point O with speed Vms-1 at an angle of elevation θ, where

tanθ = 4/3. The point O is 42.5m above the horizontal plane. The particle strikes the plane

5 seconds after it is projected.

a) Show that V = 20ms-1

b) Find the distance between O and A

Find the values of Sinθ and Cosθ first, by labelling a triangle where Tanθ = 4/3 (That is,

opposite = 4 and adjacent = 3)

V

θO

A

𝑇𝑎𝑛𝜃 =4

3

42

.5m

𝑆𝑖𝑛𝜃 =4

5

𝐶𝑜𝑠𝜃 =3

5

3

45

Opp

Adj

Hyp Use Pythagoras’ Theorem to find the missing side Remember Sinθ = opp/hyp and Cosθ = adj/hyp

Vcosθ

Vsinθ

Resolving vertically using the information given (taking upwards as positive)

The ball will be 42.5m lower after 5 seconds

𝑠 = −42.5 𝑢 = 𝑉𝑠𝑖𝑛θ 𝑣 =? 𝑎 = −9.8 𝑡 = 5

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

−42.5 = (𝑉𝑠𝑖𝑛θ)(5) +1

2(−9.8)(52)

−42.5 = 4𝑉 − 122.5

80 = 4𝑉

20 = 𝑉

Sub in values

Calculate terms (use sinθ = 4/5)

Add 122.5

Divide by 4


(𝑅 ↑)

Projectiles

6C


plane






V

θO

A

𝑇𝑎𝑛𝜃 =4

3

42

.5m

𝑆𝑖𝑛𝜃 =4

5

𝐶𝑜𝑠𝜃 =3

5

Vcosθ

Vsinθ



𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑉𝑐𝑜𝑠θ × 5

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 203

5× 5

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 60𝑚

60m

Make sure you read the question –we want the distance OA, not just the horizontal distance travelled!

Projectiles

6C


plane






V

θO

A

𝑇𝑎𝑛𝜃 =4

3

42

.5m

𝑆𝑖𝑛𝜃 =4

5

𝐶𝑜𝑠𝜃 =3

5


60m

42.52 + 602

Use Pythagoras’ Theorem to calculate the distance

= 74𝑚 (2sf)

Projectiles

6C


plane

A particle is projected from a point O with speed 35ms-1 at an angle of elevation of 30°.

The particle moves freely under gravity.

Find the length of time for which the particle is 15m or more above O

We want to find the times that the particle is exactly at 15m. There will be 2 of these, once as the particle is travelling up, and once as it is

travelling down.

15m

30°

35

35Cos30

35Sin30

Resolving vertically, taking upwards as positive

𝑠 = 15 𝑢 = 35𝑆𝑖𝑛30 𝑣 =? 𝑎 = −9.8 𝑡 =?

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

15 = (35𝑠𝑖𝑛30)(𝑡) +1

2(−9.8)(𝑡2)

15 = 35𝑡𝑠𝑖𝑛30 − 4.9𝑡2

4.9𝑡2 − 35𝑡𝑠𝑖𝑛30 + 15 = 0

𝑎 = 4.9 𝑏 = −35𝑠𝑖𝑛30 𝑐 = 15

Sub in values

Work out each term

Rearrange to a quadratic form

This will be difficult to factorise so we should use the quadratic formula – hence we need a, b and c


(𝑅 ↑)

Projectiles

6C


plane

A particle is projected from a point O with speed 35ms-1 at an angle of elevation of 30°.

The particle moves freely under gravity.

Find the length of time for which the particle is 15m or more above O

We want to find the times that the particle is exactly at 15m. There will be 2 of these, once as the particle is travelling up, and once as it is

travelling down.

15m

30°

35

35Cos30

35Sin30


𝑡 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎

𝑡 =35𝑠𝑖𝑛30 ± (−35𝑠𝑖𝑛30)2−(4 × 4.9 × 15)

2(4.9)

𝑡 = 1.43 𝑜𝑟 2.14 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 (2𝑑𝑝)

The difference between these times will be the time spent above 15m Subtract the smallest from the biggest

Time above 15m = 0.71 seconds(remember to use exact answers)

𝑎 = 4.9 𝑏 = −35𝑠𝑖𝑛30 𝑐 = 15

Projectiles

6C


vertical plane

A ball is struck by a racket at a point A which is 2m above horizontal ground.

Immediately after being struck, the ball has velocity (5i + 8j) ms-1, where i and j

are unit vectors horizontally and vertically respectively.

After being struck, the ball travels freely under gravity until is strikes the ground at

a point B, as shown. Find:

a) The greatest height above ground reached by the ball

b) The speed of the ball as it reaches B

c) The angle the velocity of the ball makes with the ground as the ball reaches B

(5i + 8j)

5i

8j

A

B

You can use the vectors in each direction as the initial velocities

Resolve vertically to find the greatest height

2m

𝑠 =? 𝑢 = 8 𝑣 = 0 𝑎 = −9.8 𝑡 =?

𝑣2 = 𝑢2 + 2𝑎𝑠

(0)2= (8)2+2(−9.8)𝑠

0 = 64 − 19.6𝑠

𝑠 = 3.3𝑚

Sub in values

Work out terms

Calculate s

Careful – the ball starts at a height of 2m, so this must be added on!

𝑠 = 5.3𝑚

𝑠 = 5.3𝑚

(𝑅 ↑)

Projectiles

6C


vertical plane









(5i + 8j)

5i

8j

A

B

2m

As the ball strikes B, its velocity will have both a horizontal and vertical component The horizontal speed is constant (5) so we do not need to

calculate this The vertical speed however will vary as the ball travels so we

need to work this out… At B the ball has travelled 2m down – resolve vertically again,

taking downwards as the positive direction

𝑠 = 2 𝑢 = −8 𝑣 =? 𝑎 = 9.8 𝑡 =?

𝑣2 = 𝑢2 + 2𝑎𝑠

𝑣2 = (−8)2+2(9.8)(2)

𝑣2 = 103.2

𝑣 = 10.2𝑚𝑠−1

Sub in values

Work out right side

Square root

So as it strikes B, the ball has a velocity of 10.2ms-1 downwards

𝑠 = 5.3𝑚

(𝑅 ↓)

Projectiles

6C


vertical plane









(5i + 8j)

5i

8j

A

B

2m

As the ball strikes B, its velocity will have both a horizontal and vertical component The horizontal speed is constant (5) so we do not need to

calculate this The vertical speed however will vary as the ball travels so we

need to work this out… At B the ball has travelled 2m down – resolve vertically again,

taking downwards as the positive direction

B

5

10.2

You can just use Pythagoras to work out the overall speed!

= 52 + 10.22

= 11𝑚𝑠−1 (2𝑠𝑓)

= 11𝑚𝑠−1

𝑠 = 5.3𝑚

(𝑅 ↓)

Projectiles

6C


vertical plane









(5i + 8j)

5i

8j

A

B

2m

We can use the same diagram to calculate the angle between the velocity and the ground…

B

5

10.2θ

𝑇𝑎𝑛𝜃 =𝑂𝑝𝑝

𝐴𝑑𝑗

𝑇𝑎𝑛𝜃 =5

10.2

𝑇𝑎𝑛𝜃 = 0.49

𝜃 = 26.2°

𝜃 = 63.8°

Sub in values

Calculate

Inverse Tan

Remember to work out the actual angle between the velocity and θ (Subtract θ from 90°)

= 63.8°

26.2°

63.8°

= 11𝑚𝑠−1

𝑠 = 5.3𝑚

Projectiles

6C

Projectiles

6D

You need to be able to derive some formulae for projectile motion

A particle is projected from a point on a horizontal plane with an initial velocity 𝑈 at an angle 𝜃 above the horizontal, and moves freely under gravity until it hits the plane at point B. Given that the acceleration due to gravity is 𝑔,

find expressions for:

a) The time of flight, 𝑇

b) The range, 𝑅, on the horizontal plane

The time of flight will be the time taken for the particle to return to its original height

ie) The displacement will be 0…

𝑈

θ

𝑈𝑐𝑜𝑠𝜃

𝑈𝑠𝑖𝑛𝜃

Resolving vertically.. (𝑅 ↑)

𝑠 = 0 𝑢 = 𝑈𝑠𝑖𝑛θ 𝑣 =? 𝑎 = −𝑔 𝑡 = 𝑇

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

0 = (𝑈𝑠𝑖𝑛𝜃)(𝑇) +1

2(−𝑔)(𝑇)2

0 = 𝑈𝑇𝑠𝑖𝑛𝜃 −1

2𝑔𝑇2

0 = 𝑇 𝑈𝑠𝑖𝑛𝜃 −1

2𝑔𝑇

Either 𝑇 = 0 (at the start of the motion), or the part inside the bracket is 0

Sub in values

Simplify

Factorise

Projectiles

6D






The time of flight will be the time taken for the particle to return to its original height

ie) The displacement will be 0…

𝑈

θ



0 = 𝑇 𝑈𝑠𝑖𝑛𝜃 −1

2𝑔𝑇

The part inside the bracket is 0

Add 1

2𝑔𝑡

0 = 𝑈𝑠𝑖𝑛𝜃 −1

2𝑔𝑇

1

2𝑔𝑇 = 𝑈𝑠𝑖𝑛𝜃

𝑔𝑇 = 2𝑈𝑠𝑖𝑛𝜃

𝑇 =2𝑈𝑠𝑖𝑛𝜃

𝑔

Multiply by 2

Divide by 𝑔


𝑔

Time of flight on a horizontal plane

Projectiles

6D






The range of flight will be the time of flight, multiplied by the horizontal speed in that

direction

We can use the result for the time of flight which we just had…

𝑈

θ




𝑔


𝑅𝑎𝑛𝑔𝑒 = 𝑈𝑐𝑜𝑠𝜃 ×2𝑈𝑠𝑖𝑛𝜃

𝑔

𝑅𝑎𝑛𝑔𝑒 = 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑠𝑝𝑒𝑒𝑑 × 𝑇𝑖𝑚𝑒 𝑜𝑓 𝑓𝑙𝑖𝑔ℎ𝑡

𝑅𝑎𝑛𝑔𝑒 =2𝑈2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

𝑔

Combine

Use 𝑆𝑖𝑛2𝜃 = 2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃

𝑅𝑎𝑛𝑔𝑒 =𝑈2𝑠𝑖𝑛2𝜃

𝑔

𝑇 =𝑈2𝑠𝑖𝑛2𝜃

𝑔

Range of flight on a horizontal plane


A particle is projected from a point with speed 𝑢 and an angle of elevation 𝜃, and moves freely under gravity. When the particle has moved a

horizontal distance 𝑥, its height above the point of projection is 𝑦.

Show that:

𝑦 = 𝑥𝑡𝑎𝑛𝜃 −𝑔𝑥2

2𝑢2(1 + 𝑡𝑎𝑛2𝜃)

u

𝜃

𝑢𝐶𝑜𝑠𝜃

𝑢𝑆𝑖𝑛𝜃

In this type of question you need to form equations for the height, 𝑦, in terms of the time, 𝑡,

as well as the horizontal distance, 𝑥, in terms of the time, 𝑡.

The equations can then be combined to eliminate 𝑡

Finding the height in terms of t Resolving vertically (upwards as positive) Use g to represent acceleration and y to represent the

height above the point of projection

𝑠 = 𝑦 𝑢 = 𝑢𝑠𝑖𝑛θ 𝑣 =? 𝑎 = −𝑔 𝑡 = 𝑡

𝑠 = 𝑢𝑡 +1

2𝑎𝑡2

𝑦 = (𝑢𝑠𝑖𝑛𝜃)(𝑡) +1

2(−𝑔)(𝑡2)

𝑦 = 𝑢𝑡𝑠𝑖𝑛𝜃 −1

2𝑔𝑡2

Sub in values from SUVAT

‘Tidy up’

We now have an expression for the height in terms of u, t and θ


2𝑔𝑡2

Projectiles

6D

(𝑅 ↑)


𝑔



𝑔





Show that:

u

𝜃

𝑢𝐶𝑜𝑠𝜃

𝑢𝑆𝑖𝑛𝜃

Finding the horizontal distance in terms of t


𝑥 = 𝑢𝑐𝑜𝑠𝜃 × 𝑡

𝑥 = 𝑢𝑡𝑐𝑜𝑠𝜃

Sub in values (remember x is to be used to represent the horizontal distance)

𝑥 = 𝑢𝑡𝑐𝑜𝑠𝜃

Projectiles

6D


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)

In this type of question you need to form equations for the height, 𝑦, in terms of the time, 𝑡,

as well as the horizontal distance, 𝑥, in terms of the time, 𝑡.

The equations can then be combined to eliminate 𝑡


2𝑔𝑡2


𝑔



𝑔





Show that:

Now we have equations for 𝑦 and 𝑥.

The final answer we want has no ‘𝑡’ terms, so this is an indication that you have to eliminate 𝑡

from the equations

The final answer is written as ‘𝑦 =‘, so it looks like we should rearrange the ‘𝑥’ equation and

substitute it into the ‘𝑦’ equation


2𝑔𝑡2 𝑥 = 𝑢𝑡𝑐𝑜𝑠𝜃

𝑥

𝑢𝑐𝑜𝑠𝜃= 𝑡

Divide by ucosθ


2𝑔𝑡2

𝑦 = 𝑢𝑥

𝑢𝑐𝑜𝑠𝜃𝑠𝑖𝑛𝜃 −

1

2𝑔

𝑥

𝑢𝑐𝑜𝑠𝜃

2

Replace t with the expression

above

𝑦 =𝑢𝑥𝑠𝑖𝑛𝜃

𝑢𝑐𝑜𝑠𝜃−

1

2𝑔

𝑥2

𝑢2𝑐𝑜𝑠2𝜃


2𝑢21

𝑐𝑜𝑠2𝜃


2𝑢2𝑠𝑒𝑐2𝜃

Group the first set of terms,

square the bracket

Cancel u’s on the first term. Sin/Cos = Tan

Group terms except for Cos on the second

1/cos = sec

𝑠𝑖𝑛2𝜃 + 𝑐𝑜𝑠2𝜃 ≡ 1

𝑡𝑎𝑛2𝜃 + 1 ≡ 𝑠𝑒𝑐2𝜃

Divide by cos2θ


2𝑢2(𝑡𝑎𝑛2𝜃 + 1)

Use the trig identity above to

replace sec

Projectiles

6D


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)


𝑔



𝑔



A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an

angle of elevation θ. The particle passes through a point B, which is 8m above the plane

and a horizontal distance of 32m from A

Find the two possible values of θ, giving your answers to the nearest degree.

(Use the formula we have just calculated)


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)

8 = 32𝑡𝑎𝑛𝜃 −9.8(32)2

2(28)2(1 + 𝑡𝑎𝑛2𝜃)

8 = 32𝑡𝑎𝑛𝜃 − 6.4(1 + 𝑡𝑎𝑛2𝜃)

8 = 32𝑡𝑎𝑛𝜃 − 6.4 + 6.4𝑡𝑎𝑛2𝜃

6.4𝑡𝑎𝑛2𝜃 + 32𝑡𝑎𝑛𝜃 − 14.4 = 0

Sub in the values we know

Calculate the large fraction

Multiply out the bracket

Rearrange to a quadratic form

𝑎 = 6.4 𝑏 = 32 𝑐 = −14.4

You can solve this like a quadratic by using the quadratic formula

Projectiles

6D


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)𝑇 =

2𝑈𝑠𝑖𝑛𝜃

𝑔



𝑔



A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an

angle of elevation θ. The particle passes through a point B, which is 8m above the plane

and a horizontal distance of 32m from A

Find the two possible values of θ, giving your answers to the nearest degree.

(Use the formula we have just calculated)

𝑎 = 6.4 𝑏 = 32 𝑐 = −14.4

𝑡𝑎𝑛𝜃 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎

6.4𝑡𝑎𝑛2𝜃 + 32𝑡𝑎𝑛𝜃 − 14.4 = 0

𝑡𝑎𝑛𝜃 =−32 ± 322 − (4 × 6.4 × −14.4)

2(6.4)

𝑡𝑎𝑛𝜃 = 0.5 𝑡𝑎𝑛𝜃 = 4.5𝑜𝑟

𝜃 = 27° 𝜃 = 77°𝑜𝑟

There will be two possibilities here:

Sub in values

Calculate each possibility

Use inverse Tan (and round

answers)

A

BIf a projectile passes through point B, there are two possible angles it could

have been launched through

Projectiles

6D


2𝑢2(1 + 𝑡𝑎𝑛2𝜃)


𝑔



𝑔


Projectiles

6D


𝑔



𝑔



2𝑢2(1 + 𝑡𝑎𝑛2𝜃)

Equation of the trajectory

𝑇 =𝑈𝑠𝑖𝑛𝜃

𝑔

Time to reach greatest height

You need to be able to derive formula from algebraic situations

You are not given these in the booklet, but can use them in problem solving if you remember them

However, it is easy to get them mixed up, so it is generally recommended that you solve non-algebraic problems using the techniques that you have already learned this

chapter

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