Projectile Motion

The trajectory of an object undergoing projectile motion consists of a vertical component and a horizontal component
- These need to be evaluated separately
Some key terms to know, and how to calculate them, are:
- Time of flight: how long the projectile is in the air
- Maximum height attained: the height at which the projectile is momentarily at rest
- Range: the horizontal distance travelled by the projectile

Projectile Motion, downloadable AS & A Level Physics revision notes

How to find the time of flight, maximum height and range

Problems involving projectile motion

There are two main considerations for solving problems involving two-dimensional motion of a projectile
- Constant velocity in the horizontal direction
- Constant acceleration in a perpendicular direction

The only force acting on the projectile, after it has been released, is the gravitational pull of the Earth on the object, or weight
There are three possible scenarios for projectile motion:
- Vertical projection
- Horizontal projection
- Projection at an angle

Worked example

To calculate vertical projection (free fall)

A science museum designed an experiment to show the fall of a feather in a vertical glass vacuum tube.

The time of fall from rest is 0.5 s.

WE - Projectile Motion Worked Example 1 question image, downloadable AS & A Level Physics revision notes

What is the length of the tube, L?

WE - Projectile Motion Worked Example 1 answer image, downloadable AS & A Level Physics revision notes

Worked example

To calculate horizontal projection

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

Worked example

To calculate projection at an angle

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

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

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

$s = - 4.3 m$

Examiner Tip

Make sure you don’t make these common mistakes:

Forgetting that deceleration is negative as the object rises
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.

Projectile Motion (DP IB Physics: SL)

Revision Note