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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 ΞΈ, based on the diagram below
3a) Given that π πππ =5
13and π is acute,
find πππ π and π‘πππ.
b) Given that π‘πππ =8
15and π is reflex,
find π πππ and πππ π.
11.5π
3.1π
π₯ = π£πππ π
v
x
y
ΞΈ
π¦ = π£π πππ
πππ π =12
13π‘πππ =
5
12
π πππ = β8
17πππ π = β
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
π = 30 π’ = 0 π£ =? π = 9.8 π‘ =?
30 = (0)(π‘) +1
2π(9.8)π‘2
π = π’π‘ +1
2ππ‘2
30 = 4.9π‘2
2.5 = π‘ (2sf)
Sub in values
Work out terms
Calculate the positive value
π‘ = 2.5
Start with a diagram!
Projectiles
6A
(π β)
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
The horizontal speed will remain constant at 20ms-1
30
m
Start with a diagram!
π»ππππ§πππ‘ππ πππ π‘ππππ = π»ππππ§πππ‘ππ π ππππ Γ π‘πππ
π»ππππ§πππ‘ππ πππ π‘ππππ = 20 Γ 2.5
π»ππππ§πππ‘ππ πππ π‘ππππ = 49π (2sf)
Make sure you use the exact value for t rather than the rounded one from part a)!
π = 49π
π‘ = 2.5
Projectiles
6A
(π β)
20ms-1
You can use the constant acceleration formulae for a projectile moving in a
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 constantβ¦
Projectiles
6A
15ms-1Start with a diagram!
You can use π¦ and π₯ to represent the vertical and horizontal displacementsβ¦
π¦
π₯
π‘ = 3π
π’π₯ = 15 π’π¦ = 0
(π β)
π = π π‘
π₯ = (15)(3)
π₯ = 45π
Sub in values β remember that π₯ is the horizontal
distance
Calculate
π₯ = 45π
The initial velocities in the horizontal and vertical directions are sometimes represented by π’π₯ and π’π¦ respectivelyβ¦
You can use the constant acceleration formulae for a projectile moving in a
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
We need to use a suvat equation to find the vertical distance, since the
velocity is not constantβ¦
Projectiles
6A
15ms-1Start with a diagram!
You can use π¦ and π₯ to represent the vertical and horizontal displacementsβ¦
π¦
π₯
π‘ = 3π
π’π₯ = 15 π’π¦ = 0
(π β)
π = π’π‘ +1
2ππ‘2
Sub in values
π₯ = 45π
The initial velocities in the horizontal and vertical directions are sometimes represented by π’π₯ and π’π¦ respectivelyβ¦
π¦ = (0)(3) +1
2(9.8)(3)2
π¦ = 44.1π
Calculate
π = π¦ π’ = 0 π£ =? π = 9.8 π‘ = 3
π¦ = 44.1π
You can use the constant acceleration formulae for a projectile moving in a
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
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
downwardsβ¦
The total distance can be found using Pythagorasβ Theoremβ¦
Projectiles
6A
15ms-1Start with a diagram!
You can use π¦ and π₯ to represent the vertical and horizontal displacementsβ¦
π¦
π₯
π‘ = 3π
π₯ = 45π
π¦ = 44.1π
45π
44.1ππ·ππ π‘ππππ
π = (45)2+(44.1)2
= 63π (2sf)
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle is projected horizontally with a speed of π ππ β1 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 constantβ¦
Therefore:
Projectiles
6A
π ms-1
Start with a diagram and label all the information you haveβ¦
90π
122.5π
π = π π‘
90 = π’π‘
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 velocityβ¦
We can use the vertical information to find when the particle hits the planeβ¦
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle is projected horizontally with a speed of π ππ β1 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.
Projectiles
6A
π ms-1
Start with a diagram and label all the information you haveβ¦
90π
122.5π
(π β)
π = 112.5 π’ = 0 π£ =? π = 9.8 π‘ =?
π = π’π‘ +1
2ππ‘2
112.5 = (0)(π‘) +1
2(9.8)π‘2
112.5 = 4.9π‘2
We can use the vertical information to find when the particle hits the planeβ¦
5 = π‘
Sub in values
Simplify right side
Calculate, and take the positive value of π‘
π = π
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle is projected horizontally with a speed of π ππ β1 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.
Projectiles
6A
π ms-1
Start with a diagram and label all the information you haveβ¦
90π
122.5π
π = π
π = π π‘
90 = π’(5)
We know that the distance travelled is 90m, after 5 seconds. The speed is π’.
(π β)
Calculate18 = π’
Now we can resolve horizontallyβ¦
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 π’, at an angle π to the horizontal,
it moves along a symmetrical curve
The speed π’ is known as the speed of projection
The angle π 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
π’ ππ β1
ΞΈ
π’ ππ β1
ΞΈ
π’πΆππ π ππ β1
π’ππππ ππ β1
Any particle projected at an angle will have an initial horizontal and vertical speed, which you will usually
need to resolve separatelyβ¦
Projectiles
6B
You need to understand how to resolve initial projection velocities that are
given at an angle (usually relative to a horizontal plane)
A particle is projected from a point on a horizontal plane with an initial velocity of 40ms-1, at an angle of πΌ above the
horizontal, where π‘πππΌ =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
π‘πππΌ =3
4
π πππΌ =3
5
πππ πΌ =4
5
πΌ
40 ππ β1
πΌ
40πππ πΌ ππ β1
40π πππΌ ππ β1
We can now use these ratio to split the initial velocity into its horizontal and vertical component parts
32 ππ β1
24 ππ β1
Calculate the exact values using the ratio we foundβ¦
32π + 24π ππ β1
You need to understand how to resolve initial projection velocities that are
given at an angle (usually relative to a horizontal plane)
A particle is projected with velocity πΌ = 3π + 5π ππ β1, where π and π 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
π
3π
5π
Initial speed β use Pythagorasβ Theoremβ¦
= (3)2+(5)2
= 34
34
Angle of projection β use tanβ¦
= πππβ15
3
= 59Β° (2sf)
So the initial projection is at speed 34 ππ β1 at an angle of 59Β° above the horizontal
59Β°
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30Β°. 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
Start with a diagram!
O A
28
30Β°
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
π =? π’ = 28π ππ30 π£ = 0 π = β9.8 π‘ =?
π£2 = π’2 + 2ππ
(0)2= (28π ππ30)2+2(β9.8)(π )
0 = 196 β 19.6π
19.6π = 196
π = 10π
Sub in values
Calculate terms
Rearrange
Calculate
10π
(π β)
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30Β°. 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
Start with a diagram!
O A
28
30Β°
28cos30
28sin30
The time of flight of the particle will be when its vertical displacement is 0 (ie β it is at the same height
it started at)
Resolve vertically and use SUVAT (again)
π = 0 π’ = 28π ππ30 π£ =? π = β9.8 π‘ =?
π = π’π‘ +1
2ππ‘2
0 = (28π ππ30)(π‘) +1
2(β9.8)(π‘2)
0 = 28π‘π ππ30 β 4.9π‘2
0 = π‘(28π ππ30 β 4.9π‘)
π‘ = 0 ππ 28π ππ30 β 4.9π‘ = 0
π‘ = 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 particleβs starting position. The value 2.9 is the time it
takes to hit the ground again
2.9π
10π
(π β)
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
vertical plane
A particle P is projected from a point O on a horizontal plane with speed 28ms-1, and with angle of elevation 30Β°. 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
Start with a diagram!
O A
28
30Β°
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)
π»ππππ§πππ‘ππ πππ π‘ππππ = π»ππππ§πππ‘ππ π ππππ Γ π‘πππ
π»ππππ§πππ‘ππ πππ π‘ππππ = 28πππ 30 Γ 2.9
π»ππππ§πππ‘ππ πππ π‘ππππ = 69π (2π π)
69π
2.9π
10π
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 ΞΈ, where
tanΞΈ = 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 SinΞΈ and CosΞΈ first, by labelling a triangle where TanΞΈ = 4/3 (That is,
opposite = 4 and adjacent = 3)
V
ΞΈO
A
ππππ =4
3
42
.5m
ππππ =4
5
πΆππ π =3
5
3
45
Opp
Adj
Hyp Use Pythagorasβ Theorem to find the missing side Remember SinΞΈ = opp/hyp and CosΞΈ = adj/hyp
VcosΞΈ
VsinΞΈ
Resolving vertically using the information given (taking upwards as positive)
The ball will be 42.5m lower after 5 seconds
π = β42.5 π’ = ππ ππΞΈ π£ =? π = β9.8 π‘ = 5
π = π’π‘ +1
2ππ‘2
β42.5 = (ππ ππΞΈ)(5) +1
2(β9.8)(52)
β42.5 = 4π β 122.5
80 = 4π
20 = π
Sub in values
Calculate terms (use sinΞΈ = 4/5)
Add 122.5
Divide by 4
Start with a diagram!
(π β)
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 ΞΈ, where
tanΞΈ = 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
V
ΞΈO
A
ππππ =4
3
42
.5m
ππππ =4
5
πΆππ π =3
5
VcosΞΈ
VsinΞΈ
Start with a diagram!
π»ππππ§πππ‘ππ πππ π‘ππππ = π»ππππ§πππ‘ππ π ππππ Γ π‘πππ
π»ππππ§πππ‘ππ πππ π‘ππππ = ππππ ΞΈ Γ 5
π»ππππ§πππ‘ππ πππ π‘ππππ = 203
5Γ 5
π»ππππ§πππ‘ππ πππ π‘ππππ = 60π
60m
Make sure you read the question βwe want the distance OA, not just the horizontal distance travelled!
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 ΞΈ, where
tanΞΈ = 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
V
ΞΈO
A
ππππ =4
3
42
.5m
ππππ =4
5
πΆππ π =3
5
Start with a diagram!
60m
42.52 + 602
Use Pythagorasβ Theorem to calculate the distance
= 74π (2sf)
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 35ms-1 at an angle of elevation of 30Β°.
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
30Β°
35
35Cos30
35Sin30
Resolving vertically, taking upwards as positive
π = 15 π’ = 35πππ30 π£ =? π = β9.8 π‘ =?
π = π’π‘ +1
2ππ‘2
15 = (35π ππ30)(π‘) +1
2(β9.8)(π‘2)
15 = 35π‘π ππ30 β 4.9π‘2
4.9π‘2 β 35π‘π ππ30 + 15 = 0
π = 4.9 π = β35π ππ30 π = 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 β hence we need a, b and c
Start with a diagram!
(π β)
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 35ms-1 at an angle of elevation of 30Β°.
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
30Β°
35
35Cos30
35Sin30
Start with a diagram!
π‘ =βπ Β± π2 β 4ππ
2π
π‘ =35π ππ30 Β± (β35π ππ30)2β(4 Γ 4.9 Γ 15)
2(4.9)
π‘ = 1.43 ππ 2.14 π ππππππ (2ππ)
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)
π = 4.9 π = β35π ππ30 π = 15
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
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
π =? π’ = 8 π£ = 0 π = β9.8 π‘ =?
π£2 = π’2 + 2ππ
(0)2= (8)2+2(β9.8)π
0 = 64 β 19.6π
π = 3.3π
Sub in values
Work out terms
Calculate s
Careful β the ball starts at a height of 2m, so this must be added on!
π = 5.3π
π = 5.3π
(π β)
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
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
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 outβ¦ At B the ball has travelled 2m down β resolve vertically again,
taking downwards as the positive direction
π = 2 π’ = β8 π£ =? π = 9.8 π‘ =?
π£2 = π’2 + 2ππ
π£2 = (β8)2+2(9.8)(2)
π£2 = 103.2
π£ = 10.2ππ β1
Sub in values
Work out right side
Square root
So as it strikes B, the ball has a velocity of 10.2ms-1 downwards
π = 5.3π
(π β)
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
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
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 outβ¦ At B the ball has travelled 2m down β 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
= 11ππ β1 (2π π)
= 11ππ β1
π = 5.3π
(π β)
Projectiles
6C
You can use the constant acceleration formulae for a projectile moving in a
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
2m
We can use the same diagram to calculate the angle between the velocity and the groundβ¦
B
5
10.2ΞΈ
ππππ =πππ
π΄ππ
ππππ =5
10.2
ππππ = 0.49
π = 26.2Β°
π = 63.8Β°
Sub in values
Calculate
Inverse Tan
Remember to work out the actual angle between the velocity and ΞΈ (Subtract ΞΈ from 90Β°)
= 63.8Β°
26.2Β°
63.8Β°
= 11ππ β1
π = 5.3π
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 π at an angle π above the horizontal, and moves freely under gravity until it hits the plane at point B. Given that the acceleration due to gravity is π,
find expressions for:
a) The time of flight, π
b) The range, π , 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 0β¦
π
ΞΈ
ππππ π
ππ πππ
Resolving vertically.. (π β)
π = 0 π’ = ππ ππΞΈ π£ =? π = βπ π‘ = π
π = π’π‘ +1
2ππ‘2
0 = (ππ πππ)(π) +1
2(βπ)(π)2
0 = πππ πππ β1
2ππ2
0 = π ππ πππ β1
2ππ
Either π = 0 (at the start of the motion), or the part inside the bracket is 0
Sub in values
Simplify
Factorise
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 π at an angle π above the horizontal, and moves freely under gravity until it hits the plane at point B. Given that the acceleration due to gravity is π,
find expressions for:
a) The time of flight, π
b) The range, π , 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 0β¦
π
ΞΈ
ππππ π
ππ πππ
0 = π ππ πππ β1
2ππ
The part inside the bracket is 0
Add 1
2ππ‘
0 = ππ πππ β1
2ππ
1
2ππ = ππ πππ
ππ = 2ππ πππ
π =2ππ πππ
π
Multiply by 2
Divide by π
π =2ππ πππ
π
Time of flight on a horizontal plane
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 π at an angle π above the horizontal, and moves freely under gravity until it hits the plane at point B. Given that the acceleration due to gravity is π,
find expressions for:
a) The time of flight, π
b) The range, π , on the horizontal plane
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 hadβ¦
π
ΞΈ
ππππ π
ππ πππ
π =2ππ πππ
π
Time of flight on a horizontal plane
π ππππ = ππππ π Γ2ππ πππ
π
π ππππ = π»ππππ§πππ‘ππ π ππππ Γ ππππ ππ ππππβπ‘
π ππππ =2π2π ππππππ π
π
Combine
Use πππ2π = 2π ππππππ π
π ππππ =π2π ππ2π
π
π =π2π ππ2π
π
Range of flight on a horizontal plane
You need to be able to derive some formulae for projectile motion
A particle is projected from a point with speed π’ and an angle of elevation π, and moves freely under gravity. When the particle has moved a
horizontal distance π₯, its height above the point of projection is π¦.
Show that:
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
u
π
π’πΆππ π
π’ππππ
In this type of question you need to form equations for the height, π¦, in terms of the time, π‘,
as well as the horizontal distance, π₯, in terms of the time, π‘.
The equations can then be combined to eliminate π‘
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
π = π¦ π’ = π’π ππΞΈ π£ =? π = βπ π‘ = π‘
π = π’π‘ +1
2ππ‘2
π¦ = (π’π πππ)(π‘) +1
2(βπ)(π‘2)
π¦ = π’π‘π πππ β1
2ππ‘2
Sub in values from SUVAT
βTidy upβ
We now have an expression for the height in terms of u, t and ΞΈ
π¦ = π’π‘π πππ β1
2ππ‘2
Projectiles
6D
(π β)
π =2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
You need to be able to derive some formulae for projectile motion
A particle is projected from a point with speed π’ and an angle of elevation π, and moves freely under gravity. When the particle has moved a
horizontal distance π₯, its height above the point of projection is π¦.
Show that:
u
π
π’πΆππ π
π’ππππ
Finding the horizontal distance in terms of t
π»ππππ§πππ‘ππ πππ π‘ππππ = π»ππππ§πππ‘ππ π ππππ Γ π‘πππ
π₯ = π’πππ π Γ π‘
π₯ = π’π‘πππ π
Sub in values (remember x is to be used to represent the horizontal distance)
π₯ = π’π‘πππ π
Projectiles
6D
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
In this type of question you need to form equations for the height, π¦, in terms of the time, π‘,
as well as the horizontal distance, π₯, in terms of the time, π‘.
The equations can then be combined to eliminate π‘
π¦ = π’π‘π πππ β1
2ππ‘2
π =2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
You need to be able to derive some formulae for projectile motion
A particle is projected from a point with speed π’ and an angle of elevation π, and moves freely under gravity. When the particle has moved a
horizontal distance π₯, its height above the point of projection is π¦.
Show that:
Now we have equations for π¦ and π₯.
The final answer we want has no βπ‘β terms, so this is an indication that you have to eliminate π‘
from the equations
The final answer is written as βπ¦ =β, so it looks like we should rearrange the βπ₯β equation and
substitute it into the βπ¦β equation
π¦ = π’π‘π πππ β1
2ππ‘2 π₯ = π’π‘πππ π
π₯
π’πππ π= π‘
Divide by ucosΞΈ
π¦ = π’π‘π πππ β1
2ππ‘2
π¦ = π’π₯
π’πππ ππ πππ β
1
2π
π₯
π’πππ π
2
Replace t with the expression
above
π¦ =π’π₯π πππ
π’πππ πβ
1
2π
π₯2
π’2πππ 2π
π¦ = π₯π‘πππ βππ₯2
2π’21
πππ 2π
π¦ = π₯π‘πππ βππ₯2
2π’2π ππ2π
Group the first set of terms,
square the bracket
Cancel uβs on the first term. Sin/Cos = Tan
Group terms except for Cos on the second
1/cos = sec
π ππ2π + πππ 2π β‘ 1
π‘ππ2π + 1 β‘ π ππ2π
Divide by cos2ΞΈ
π¦ = π₯π‘πππ βππ₯2
2π’2(π‘ππ2π + 1)
Use the trig identity above to
replace sec
Projectiles
6D
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
π =2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
You need to be able to derive some formulae for projectile motion
A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an
angle of elevation ΞΈ. 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 ΞΈ, giving your answers to the nearest degree.
(Use the formula we have just calculated)
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
8 = 32π‘πππ β9.8(32)2
2(28)2(1 + π‘ππ2π)
8 = 32π‘πππ β 6.4(1 + π‘ππ2π)
8 = 32π‘πππ β 6.4 + 6.4π‘ππ2π
6.4π‘ππ2π + 32π‘πππ β 14.4 = 0
Sub in the values we know
Calculate the large fraction
Multiply out the bracket
Rearrange to a quadratic form
π = 6.4 π = 32 π = β14.4
You can solve this like a quadratic by using the quadratic formula
Projectiles
6D
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)π =
2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
You need to be able to derive some formulae for projectile motion
A particle is projected from a point A on a horizontal plane, with initial speed 28ms-1 and an
angle of elevation ΞΈ. 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 ΞΈ, giving your answers to the nearest degree.
(Use the formula we have just calculated)
π = 6.4 π = 32 π = β14.4
π‘πππ =βπ Β± π2 β 4ππ
2π
6.4π‘ππ2π + 32π‘πππ β 14.4 = 0
π‘πππ =β32 Β± 322 β (4 Γ 6.4 Γ β14.4)
2(6.4)
π‘πππ = 0.5 π‘πππ = 4.5ππ
π = 27Β° π = 77Β°ππ
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
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
π =2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
Projectiles
6D
π =2ππ πππ
π
Time of flight on a horizontal plane
π =π2π ππ2π
π
Range of flight on a horizontal plane
π¦ = π₯π‘πππ βππ₯2
2π’2(1 + π‘ππ2π)
Equation of the trajectory
π =ππ πππ
π
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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