Projectile Motion

What is a projectile?

A projectile is a particle moving freely (non-powered), under gravity, in a two-dimensional plane
Examples of projectile motion include throwing a ball, jumping off a diving board and hitting a baseball with a baseball bat
In these examples, it is assumed that:
- Resistance from the air or liquid (known as fluid resistance) the object is travelling through is negligible
- Acceleration due to free-fall, g is constant as the object is moving close to the surface of the Earth

2-6-1-horizontal---vertical-components-diagram-1

Examples of objects in a projectile motion trajectory

An object is sent into a projectile motion trajectory with a resultant velocity, u at an angle, θ to the horizontal
- Examples of this include a ball thrown from a height and a cannonball launched from a cannon

4-3-4-resultant-velocity-angle-of-projection

An object in a projectile motion trajectory has a resultant velocity at a given angle to the horizontal ground

Some key terms to know, and how to calculate them, are:
- Time of flight (total time): how long the projectile is in the air.
  - For typical projectile motion, the time to the maximum height is half of the total time
- Maximum height attained: the height at which the projectile is momentarily at rest
  - This is when the vertical velocity component = 0
  - When the projectile is released and lands on the ground the projectile is at its maximum height when half of its total time has elapsed
- Range: the horizontal distance travelled by the projectile

4-3-4-max-height

An object in projectile motion will have a vertical velocity of zero at maximum height when half the time has elapsed

Horizontal and Vertical Components

The trajectory of an object undergoing projectile motion consists of a vertical component and a horizontal component
- These quantities are independent of each other
- Displacement, velocity and acceleration are all vector quantities that are different in both components
- They need to be evaluated separately using the SUVAT Equations

	Horizontal Component	Vertical Component
Displacement	Maximum range at the end of the motion when the total time has elapsed Half the range at the maximum height when half the time has elapsed	Maximum height is at the top of the motion when half the time has elapsed
Velocity	Constant	Zero at maximum height
Acceleration	Zero (because velocity remains constant)	Acceleration of free fall, g = 9.8 ms⁻² Positive when an object is falling towards Earth Negative when an object is moving away from Earth

Acceleration and horizontal velocity are always constant whilst vertical velocity changes

The resultant velocity of an object in projectile motion can be split into its horizontal and vertical vector components using trigonometry where:
- Vertical component = opposite side of the projectile triangle
  - opposite = sinθ × hyp = u sinθ
- Horizontal component = adjacent side of the projectile triangle
  - adjacent = cosθ × hyp = u cosθ

The resultant velocity at an angle to the horizontal can be resolved using trigonometry into the horizontal and vertical components

It can be helpful to see how different equations calculate different quantities using SUVAT equations
Examples of obtaining the equations for total time, maximum height and range are shown below

Projectile Motion, downloadable AS & A Level Physics revision notes

Examples of using SUVAT equations to determine the time of flight, maximum height and range of a projectile

Solving problems with projectiles

You may be required to calculate the missing quantities from the following projectile motion scenarios:
- Vertical projection above the horizontal
- Vertical projection below the horizontal
- Horizontal projection
- Projection at an angle, the most common scenario

Worked example

A stone is dropped from the top of a cliff 50.0 m high at an angle of 25.0° below the horizontal. The stone has an initial speed of 30.0 ms⁻¹and follows a curved trajectory. The stone hits the ground at a horizontal distance D from the base of the cliff with a vertical velocity of 21.2 ms⁻¹ .

we-below-vertical-angle

Calculate the distance D.

Answer:

Step 1: Understand the information given in the question

The final vertical velocity of the stone, v = 21.2 ms⁻¹
The horizontal velocity will remain constant throughout the motion
The question wants us to calculate the range of the stone
The vertical acceleration is +g ms⁻²

Step 2: Resolve velocity into the vertical and horizontal components

Draw a triangle on the diagram to show the vertical and horizontal velocity components

we-method-step-trig-below-horizontal

Calculate the initial vertical component of velocity, u_v using trigonometry:

- u_v = opposite side
- u_v = sinθ × hypotenuse side
- u_v = sin(25) × 30 = 12.68 ms⁻¹
Calculate the initial horizontal component of velocity, u_H using trigonometry:
- u_H = adjacent side
- u_H = cosθ × hypotenuse side
- u_H= cos(25) × 30 = 27.19 ms⁻¹

Step 3: Consider the equations of motion in the vertical and horizontal directions

	Vertical Motion	Horizontal Motion
u	12.68 ms⁻¹	27.19 ms⁻¹
v	21.2 ms⁻¹	27.19 ms⁻¹
a	+9.81 ms⁻²	0 ms⁻²
t
s	50 m	D m ?

Step 4: Calculate the time of flight from the vertical motion

v = u + at

21.2 = −12.68 + 9.81t

21.2 + 12.68 = 9.81t

$t = \frac{33.88}{9.81}$

t = 3.45 s

Step 5: Calculate the range of the stone, D using the elapsed time and horizontal motion

s = ut

D = 27.19 × 3.45

D = 93.8 m = 94 m (2 s.f.)

Worked example

A ball is thrown from a point P with an initial velocity u of 12 m s^-1 at 50° to the horizontal.

What is the value of the maximum height at Q? (ignoring air resistance)

WE - Projectile Motion Worked Example 3 question image, downloadable AS & A Level Physics revision notes

Answer:

Step 1: Consider the situation

In this question, vertical motion only needs to be considered to find the vertical height

Step 2: List the known quantities

u = 12sin(50) ms⁻¹
v = 0 ms⁻¹
a = −9.81 ms⁻²
s = ?

Step 3: State the correct kinematic equation

$v^{2} = u^{2} + 2 a s$

Step 4: Rearrange the equation to make height, s the subject

$\frac{v^{2} - u^{2}}{2 a} = s$

Step 5: Substitute in the known quantities and calculate maximum height, s

$s = \frac{0^{2} - {(12 \sin (50))}^{2}}{2 \times - 9.81}$

$s = - 4.3 m$

Worked example

A motorcycle stunt-rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.

What was the speed at take-off? (ignoring air resistance)

WE - Projectile Motion Worked Example 2 question image, downloadable AS & A Level Physics revision notes

WE - Projectile Motion Worked Example 2 answer image, downloadable AS & A Level Physics revision notes

Exam Tip

Make sure you don’t make these common mistakes:

Mixing up positive and negative values for vectors
Mixing up velocities and distances between horizontal and vertical motion
Confusing the direction of sin θ and cos θ
Not converting units (mm, cm, km etc.) to metres

Further, it is worth noting that projectile motion is typically symmetrical when air resistance is ignored allowing for use of the peak to find the time of total flight or total horizontal distance by doubling the amount to get from the start point to the peak.

In these exam questions, unless specified, fluid resistance can be ignored

Projectile Motion (SL IB Physics)

Revision Note