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HSC Mathematics Extension 1

Applications of Calculus to the Physical World

Week 5

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HSC Mathematics Extension 1 2 Applications of Calculus to the Physical

World

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Week 5 – Theory Projectile Motion:

Projectile motion is the path that a thrown or projected particle will take under the force of gravity

and neglecting all other forms of friction such as air resistance.

A particle is projected at an angle 𝜶 known as the angle of projection with an initial velocity, 𝑽. The

trajectory is the path of the projected particle and it can be described using parametric equations or

Cartesian equation. The range is the maximum horizontal distance between the initial position and

the final position of the particle. The maximum height is the maximum vertical distance between

particle and the horizontal plane. These definitions are illustrated in the diagram below.

Since a projectile is a two-dimensional motion, we can describe the motion in terms of the

horizontal and vertical components.

Acceleration:

The particle is only subjected to acceleration due to gravity and it is acting directly vertically

downwards. Hence, the acceleration in the horizontal direction is zero and the acceleration in the

vertical direction is −𝑔, where 𝑔 is the acceleration due to gravity. Commonly, we assume 𝑔

as 10 m/s2 or 9.8 m/s2.


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Velocity:

The particle is projected at an angle 𝛼 with an initial velocity 𝑉. So by sketching the following

triangle, we can express the velocity in terms of horizontal and vertical components.

Horizontal Component

Vertical Component

Acceleration:

∴ �̈� = 0 Velocity:

�̇� = ∫ �̈� . 𝑑𝑡

= ∫ 0 . 𝑑𝑡

�̇� = 𝐶1 When 𝑡 = 0, �̇� = 𝑉 cos 𝛼, so

∴ �̇� = 𝑉 cos 𝛼 Displacement:

𝑥 = ∫ �̇� . 𝑑𝑡

= ∫ 𝑉 cos 𝛼 . 𝑑𝑡

𝑥 = 𝑉 cos 𝛼 𝑡 + 𝐶2

Acceleration:

∴ �̈� = −𝑔 Velocity:

�̇� = ∫ �̈� . 𝑑𝑡

= ∫ −𝑔 . 𝑑𝑡

�̇� = −𝑔𝑡 + 𝐷1 When 𝑡 = 0, �̇� = 𝑉 sin 𝛼, so

𝑉 sin 𝛼 = −𝑔(0) + 𝐷1 𝐷1 = 𝑉 sin 𝛼

∴ �̇� = 𝑉 sin 𝛼 − 𝑔𝑡 Displacement:

𝑦 = ∫ �̇� . 𝑑𝑡

= ∫ 𝑉 sin 𝛼 − 𝑔𝑡 . 𝑑𝑡

𝑦 = 𝑉 sin 𝛼 𝑡 −1

2𝑔𝑡2 + 𝐷2


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When 𝑡 = 0, 𝑥 = 0,

0 = 𝑉 cos 𝛼 (0) + 𝐶2 𝐶2 = 0

∴ 𝑥 = 𝑉 cos 𝛼 𝑡

When 𝑡 = 0, 𝑦 = 0,

0 = 𝑉 sin 𝛼 (0) −1

2𝑔(0)2 + 𝐷2

𝐷2 = 0

∴ 𝑦 = 𝑉 sin 𝛼 𝑡 −1

2𝑔𝑡2

NOTE: students need to learn to derive the above six equations from first principles, unless the

equations are given in the question.

The Equation of the Trajectory:

The equations 𝑥 = 𝑉 cos 𝛼 𝑡 and 𝑦 = 𝑉 sin 𝛼 𝑡 −1

2𝑔𝑡2 are the parametric equations of the

trajectory, where 𝑡 is the parameter. The Cartesian equation can be found by eliminating 𝑡 from the

equations.

𝑥 = 𝑉 cos 𝛼 𝑡

So,

𝑡 =𝑥

𝑉 cos 𝛼− − − −(1)

And,


2𝑔𝑡2 − − − −(2)

Substitute (1) into (2),

𝑦 = 𝑉 sin 𝛼 (𝑥

𝑉 cos 𝛼) −

1

2𝑔 (

𝑥

𝑉 cos 𝛼)

2

𝑦 = 𝑥 tan 𝛼 −𝑔

2(

𝑥2

𝑉2 cos2 𝛼)

𝑦 = 𝑥 tan 𝛼 −𝑔𝑥2

2𝑉2(sec2 𝛼)

Since sec2 𝛼 = 1 + tan2 𝛼, so

∴ 𝑦 = 𝑥 tan 𝛼 −𝑔𝑥2

2𝑉2(1 + tan2 𝛼)

This is the Cartesian equation of the trajectory.


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Time of Flight and Range:

Time of flight is the time required for the projected particle to finish its trajectory and it occurs when

the particle strikes the horizontal plane, i.e. when 𝑦 = 0.

Since,


2𝑔𝑡2

Substituting 𝑦 = 0,

𝑉 sin 𝛼 𝑡 −1

2𝑔𝑡2 = 0

𝑡 (𝑉 sin 𝛼 −1

2𝑔𝑡) = 0

That is,

𝑡 = 0 or 𝑉𝑠𝑖𝑛 𝛼 −1

2𝑔𝑡 = 0

When,

𝑉 sin 𝛼 −1

2𝑔𝑡 = 0

𝑉 sin 𝛼 =1

2𝑔𝑡

𝑡 =2𝑉 sin 𝛼

𝑔

But 𝑡 = 0 is the initial time, therefore the time of flight is

∴ 𝑡 =2𝑉 sin 𝛼

𝑔

As mentioned before, the range is the maximum horizontal distance between the initial position and

the final position of the particle. Hence, the range is the 𝑥 position of the particle at the time of

flight.

Substitute 𝑡 =2𝑉 sin 𝛼

𝑔,

𝑥 = 𝑉 cos 𝛼 𝑡

𝑥 = 𝑉 cos 𝛼 (2𝑉 sin 𝛼

𝑔)


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𝑥 =2𝑉2 sin 𝛼 cos 𝛼

𝑔

Since 2 sin 𝛼 cos 𝛼 = sin 2𝛼, therefore the range is

∴ 𝑥 =𝑉2 sin 2𝛼

𝑔

It can be seen from the above equation that the maximum range is 𝑥 =𝑉2

𝑔 and it occurs when 𝛼 =

45°. This is because 𝑥 =𝑉2 sin 2𝛼

𝑔 is maximum when sin 2𝛼 = 1, i.e. when 𝛼 = 45°.

Maximum Height:

The particle reaches its maximum height at its stationary point, i.e. when �̇� = 0.

Since,

�̇� = 𝑉 sin 𝛼 − 𝑔𝑡

Substitute �̇� = 0,

𝑉 sin 𝛼 − 𝑔𝑡 = 0

𝑡 =𝑉 sin 𝛼

𝑔

This is the time when the maximum height occurs.

Since,


2𝑔𝑡2

Substitute 𝑡 =𝑉 sin 𝛼

𝑔 to find the maximum height,

𝑦 = 𝑉 sin 𝛼 (𝑉 sin 𝛼

𝑔) −

1

2𝑔 (

𝑉 sin 𝛼

𝑔)

2

𝑦 =𝑉2 sin2 𝛼

𝑔−

𝑔

2(

𝑉2 sin2 𝛼

𝑔2 )

𝑦 =𝑉2 sin2 𝛼

𝑔−

𝑉2 sin2 𝛼

2𝑔


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𝑦 =2𝑉2 sin2 𝛼

2𝑔−

𝑉2 sin2 𝛼

2𝑔

Therefore the maximum height is:

∴ 𝑦 =𝑉2 sin2 𝛼

2𝑔

Example:

A soccer player kicks a ball on the ground with initial velocity 30 m/s at an angle of 30°. Assuming

the acceleration due to gravity is 10 m/s2. Find:

(i) The Cartesian equation of the trajectory.

(ii) The time of flight and the range.

(iii) The maximum height and the time when it occurs.

(iv) The speed and angle when the ball strikes the ground.

Solution:

(i)

Horizontal Motion Vertical Motion

∴ �̈� = 0

�̇� = ∫ 0 . 𝑑𝑡

�̇� = 𝐶1 When 𝑡 = 0, �̇� = 𝑉 cos 𝛼 �̇� = 30 cos 30°

�̇� = 15√3

∴ �̇� = 15√3

𝑥 = ∫ 15√3 . 𝑑𝑡

𝑥 = 15√3𝑡 + 𝐶2 When 𝑡 = 0, 𝑥 = 0,

0 = 15√3(0) + 𝐶2 𝐶2 = 0

∴ 𝑥 = 15√3𝑡

∴ �̈� = −10

�̇� = ∫ −10 . 𝑑𝑡

�̇� = −10𝑡 + 𝐷1 When 𝑡 = 0, �̇� = 𝑉 sin 𝛼 �̇� = 30 sin 30° �̇� = 15

15 = −10(0) + 𝐷1 𝐷1 = 15

∴ �̇� = 15 − 10𝑡

𝑦 = ∫ 15 − 10𝑡 . 𝑑𝑡

𝑦 = 15𝑡 − 5𝑡2 + 𝐷2 When 𝑡 = 0, 𝑦 = 0,

0 = 15(0) − 5(0)2 + 𝐷2 𝐷2 = 0

∴ 𝑦 = 15𝑡 − 5𝑡2


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Substitute 𝑡 =𝑥

15√3 into 𝑦 = 15𝑡 − 5𝑡2,

𝑦 = 15 (𝑥

15√3) − 5 (

𝑥

15√3)

2

=𝑥

√3− 5 (

𝑥2

675)

∴ 𝑦 =𝑥

√3−

𝑥2

135, which is the Cartesian equation of the trajectory

(ii) When 𝑦 = 0,

15𝑡 − 5𝑡2 = 0

5𝑡(3 − 𝑡) = 0

𝑡 = 0, 3

Since 𝑡 = 0 is the starting point so 𝑡 = 3 is the time of flight.

When 𝑡 = 3,

𝑥 = 15√3𝑡

𝑥 = 15√3(3)

∴ 𝑥 = 45√3 m

∴ The range is 45√3 metres.

(iii) When �̇� = 0,

15 − 10𝑡 = 0

10𝑡 = 15

𝑡 = 1.5 seconds

When 𝑡 = 1.5,

𝑦 = 15𝑡 − 5𝑡2

𝑦 = 15(1.5) − 5(1.5)2

𝑦 = 11.25 m

∴ The maximum height is 11.25 metres and it occurs after 1.5 seconds.

(iv) When 𝑡 = 3,

�̇� = 15√3 and

�̇� = 15 − 10𝑡

�̇� = 15 − 10(3)

�̇� = −15


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By Pythagoras’ Theorem,

𝑉𝑓𝑖𝑛𝑎𝑙 = √(−15)2 + (15√3)2

∴ 𝑉𝑓𝑖𝑛𝑎𝑙 = 30 m/s

tan 𝛼 = −15

15√3

tan 𝛼 = −1

√3

∴ 𝛼 = −30°

∴ The speed is 30 m/s and the angle is −30° when the ball strikes the ground.

Example:

A tennis ball is thrown from the top of a 30 m building at an angle 𝛼 to the ground level

where tan 𝛼 =4

3 with an initial velocity of 20 m/s. Neglecting air resistance and assuming the

acceleration due to gravity is 10 m/s2, find:

(i) The Cartesian equation of the path.

(ii) The maximum height reached and the time when it occurs.

Solution:

(i)


∴ �̈� = 0

�̇� = ∫ 0 . 𝑑𝑡

�̇� = 𝐶1 When 𝑡 = 0, �̇� = 𝑉 cos 𝛼

∴ �̈� = −10

�̇� = ∫ −10 . 𝑑𝑡

�̇� = −10𝑡 + 𝐷1 When 𝑡 = 0, �̇� = 𝑉 sin 𝛼

�̇� = 20 (4

5)

�̇� = 16

16 = −10(0) + 𝐷1


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�̇� = 20 (3

5)

�̇� = 12

∴ �̇� = 12

𝑥 = ∫ 12 . 𝑑𝑡

𝑥 = 12𝑡 + 𝐶2 When 𝑡 = 0, 𝑥 = 0,

0 = 12(0) + 𝐶2 𝐶2 = 0

∴ 𝑥 = 12𝑡

𝐷1 = 16

∴ �̇� = 16 − 10𝑡

𝑦 = ∫ 16 − 10𝑡 . 𝑑𝑡

𝑦 = 16𝑡 − 5𝑡2 + 𝐷2 When 𝑡 = 0, 𝑦 = 30,

30 = 15(0) − 5(0)2 + 𝐷2 𝐷2 = 30

∴ 𝑦 = 30 + 16𝑡 − 5𝑡2

Substitute 𝑡 =𝑥

12 into 𝑦 = 30 + 16𝑡 − 5𝑡2,

𝑦 = 30 + 16 (𝑥

12) − 5 (

𝑥

12)

2

∴ 𝑦 = 30 +4

3𝑥 −

5

144𝑥2

(ii) When �̇� = 0,

16 − 10𝑡 = 0

10𝑡 = 16

𝑡 = 1.6 seconds

When 𝑡 = 1.6,

𝑦 = 30 + 16(1.6) − 5(1.6)2

𝑦 = 42.8 metres

∴ The maximum height is 42.8 metres and it occurs after 1.6 seconds.


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Example:

A ball is thrown so that it just clears a wall that is 25 metres horizontally and 13 metres vertically

from the point of projection. If its range is 45 metres, find the speed and angle of projection.

Solution:


∴ �̈� = 0

�̇� = ∫ 0 . 𝑑𝑡

�̇� = 𝐶1 When 𝑡 = 0, �̇� = 𝑉 cos 𝛼,

𝐶1 = 𝑉 cos 𝛼

∴ �̇� = 𝑉 cos 𝛼

𝑥 = ∫ 𝑉 cos 𝛼 . 𝑑𝑡

𝑥 = 𝑉 cos 𝛼 𝑡 + 𝐶2 When 𝑡 = 0, 𝑥 = 0,

0 = 𝑉 cos 𝛼 (0) + 𝐶2 𝐶2 = 0

∴ 𝑥 = 𝑉 cos 𝛼 𝑡

∴ �̈� = −10

�̇� = ∫ −10 . 𝑑𝑡

�̇� = −10𝑡 + 𝐷1 When 𝑡 = 0, �̇� = 𝑉 sin 𝛼,

𝑉 sin 𝛼 = −10(0) + 𝐷1 𝐷1 = 𝑉 sin 𝛼

∴ �̇� = 𝑉 sin 𝛼 − 10𝑡

𝑦 = ∫ 𝑉 sin 𝛼 − 10𝑡 . 𝑑𝑡

𝑦 = 𝑉 sin 𝛼 𝑡 − 5𝑡2 + 𝐷2 When 𝑡 = 0, 𝑦 = 0,

0 = 𝑉 sin 𝛼 (0) − 5(0)2 + 𝐷2 𝐷2 = 0

∴ 𝑦 = 𝑉 sin 𝛼 𝑡 − 5𝑡2

The Cartesian equation of the trajectory can be found by substituting 𝑡 =𝑥

𝑉 cos 𝛼 into 𝑦 = 𝑉 sin 𝛼 𝑡 −

5𝑡2.

𝑦 = 𝑉 sin 𝛼 (𝑥

𝑉 cos 𝛼) − 5 (

𝑥

𝑉 cos 𝛼)

2

𝑦 = 𝑥 tan 𝛼 −5𝑥2

𝑉2sec2 𝛼

Given that when 𝑥 = 45, 𝑦 = 0 and 𝑥 = 25, 𝑦 = 13,


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So,

45 tan 𝛼 −5(45)2

𝑉2sec2 𝛼 = 0

45 tan 𝛼 −10125

𝑉2sec2 𝛼 = 0

10125 sec2 𝛼

𝑉2= 45 tan 𝛼

𝑉2 =10125 sec2 𝛼

45 tan 𝛼

𝑉2 =225 sec2 𝛼

tan 𝛼− − − −(1)

And,

13 = 25 tan 𝛼 −5(25)2

𝑉2sec2 𝛼

13 = 25 tan 𝛼 −3125 sec2 𝛼

𝑉2

3125 sec2 𝛼

𝑉2= 25 tan 𝛼 − 13

𝑉2 =3125 sec2 𝛼

25 tan 𝛼 − 13− − − −(2)

Equate equations (1) and (2),

225 sec2 𝛼

tan 𝛼=

3125 sec2 𝛼

25 tan 𝛼 − 13

225(25 tan 𝛼 − 13) = 3125 tan 𝛼

5625 tan 𝛼 − 2925 = 3125 tan 𝛼

2500 tan 𝛼 = 2925

tan 𝛼 =117

100

∴ 𝛼 ≈ 49°29′


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When 𝛼 = 49°29′,

𝑉2 =225 sec2 𝛼

tan 𝛼

=225 sec2(49°29′)

tan(49°29′)

= 45529

52

∴ 𝑉 ≈ 21.34 m/s


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Week 5 – Homework Projectile Motion:

1. A particle is projected with initial velocity 50 m/s at an angle of inclination of 45° from a

horizontal plane. Assuming the acceleration due to gravity is 10 m/s2.

(i) Derive the Cartesian equation of the path of projection.

(ii) Find the maximum height and the time when it occurs.

(iii) Find the time of flight and the range.

(iv) Find the speed and angle when the particle strikes the horizontal plane.

2. Find the velocity 𝑣 and the angle of projection 𝛼 of a projected particle if given

(i) �̇� = 12 m/s and �̇� = 5 m/s.

(ii) �̇� = 4 m/s and �̇� = −3 m/s.


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3. A ball is thrown from the top of a 40 m building with an initial speed of 20 m/s at an angle

of 30°. Assuming the acceleration due to gravity is 10 m/s2, find:

(i) The Cartesian equation of the trajectory.

(ii) The maximum height reached and its speed then.

(iii) Find the time of flight and the range.


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4. A rock is thrown from the top of a 50 m cliff at an angle 𝜃 to the ground level where tan 𝜃 =5

12

with an initial speed of 39 m/s. Assuming gravitational acceleration is 10 m/s2, find:

(i) The parametric equations of the path of trajectory and hence the Cartesian equation.

(ii) The maximum height and the time when it occurs.

(iii) The flight time and the range.

(iv) The angle and speed when the rock strikes the ground.


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5. A stone is projected horizontally from the top of a lighthouse 125 m above sea level with an

initial velocity of 30 m/s. Assuming gravitational acceleration is 10 m/s2, find:

(i) How long it takes to reach the water.

(ii) How far is the stone when it strikes the water.

(iii) The speed and angle when the stone strikes the water.


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6. A tennis ball is projected at ground level. After 2 seconds, it just clears a net 2.5 metres high at a

horizontal distance of 30 metres. Find:

(i) The angle of projection and the initial velocity.

(ii) The time of flight and the range.


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7. A bullet was fired to hit a target which is 200 m horizontally from the starting point and 25 m

high. If the bullet was fired from the ground level with an initial speed of 200 m/s, find the two

possible angles of elevation which the gun can be set to hit the target.


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8. A stone is thrown so that it just clears a wall that is 45 metres horizontally and 22.5 metres

vertically from the projection point. If its range is 90 metres, find:

(i) The angle and speed of projection.

(ii) The time of flight.

(iii) The speed and angle when it strikes the ground.


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9. A ball is thrown from the top of a 70 m building with an initial speed of 60 m/s to hit a

goal 240 m horizontally and 70 m vertically below the point of projection. Find:

(i) The angles of projection.

(ii) The speeds and angles when it hits the goal.


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10. A particle is projected and at any time 𝑡, the parametric equations of its trajectory are 𝑥 = 60𝑡

and 𝑦 = 12𝑡 − 5𝑡2. Find:

(i) The angle of projection.

(ii) The initial velocity.

(iii) The maximum height and the time taken to reach that height.

(iv) The time of flight and the range.


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11. Find the velocity of projection of a particle if its angle of projection is 15° and its range is 160

metres.


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12. A ball was dropped from a helicopter travelling at 300 km/h at a height of 600 metres. Find:

(i) The time it takes the ball to reach the ground.

(ii) The distance the ball travelled horizontally from the point of release.


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13. A particle is projected and its velocity in the horizontal and vertical components are �̇� = 5 m/s

and �̇� = 8 m/s respectively. Find the time of flight and the range.


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14. A particle is projected to just clear two walls of height 10 metres and the two walls are 5 metres

and 8 metres from the point of projection. Prove that if 𝜃 is the angle of projection then tan 𝜃 =13

4.


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15. A particle is projected to just clear two walls of height ℎ metres and the two walls are 𝑎 metres

and 𝑏 metres from the point of projection. Prove that if 𝜃 is the angle of projection then tan 𝜃 =ℎ(𝑎+𝑏)

𝑎𝑏.


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16.

(i) Show that the maximum height of a particle 𝑃 projected from the origin with an initial

velocity of 𝑉 at an angle of 𝛼 is 𝑉2 sin2 𝛼

2𝑔, where 𝑔 is the gravitational acceleration.

(ii) A second particle 𝑄 is also projected from the origin but with an initial velocity of 5

3𝑉 at an

angle of 1

2𝛼. If the two particles 𝑃 and 𝑄 both reach the same maximum height, prove

that 𝛼 = cos−1 7

18.

End of homework

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