Projectile Motion

The trajectory of an object undergoing projectile motion consists of a vertical component and a horizontal component
- The vertical and horizontal components of the motion are completely independent of one another
- Therefore, they 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 calculation example

Projectile Motion, downloadable AS & A Level Physics revision notes

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

Remember: the only force acting on the projectile, after it has been released, is gravity
There are three possible scenarios for projectile motion:
- Vertical projection
- Horizontal projection
- Projection at an angle
Let’s consider each in turn:

Worked example

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?

Answer:

Step 1: List the known quantities

Acceleration, a = 9.81 m s^–2
Initial velocity, u = 0
Time, t = 0.5 s
Displacement, s = L = ?

Step 2: State the relevant equation of motion

The equation linking a, u, t and s is

$s = u t + \frac{1}{2} a t^{2}$

Step 3: Substitute in the values

$L = \frac{1}{2} g t^{2}$

$L = \frac{1}{2} \times 9.81 \times {(0.5)}^{2} = 1.2 m$

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.

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

What was the speed at take-off?

Answer:

Step 1: List the known quantities for the vertical motion

Displacement, s = 1.25 m
Acceleration, a = 9.81 m s^–2
Initial velocity, u = 0
Time, t = ?

Step 2: State the relevant equation for the vertical motion

$s = u t + \frac{1}{2} a t^{2}$

Step 3: Rearrange for the time, t

$s = \frac{1}{2} g t^{2}$

$t = \sqrt{\frac{2 s}{g}}$

Step 4: Substitute the values

$t = \sqrt{\frac{2 \times 1.25}{9.81}} = 0.5 s$

Step 5: List the known quantities for the horizontal motion

Displacement, s = 10 m
Acceleration, a = 0
Time, t = 0.5 s
Initial velocity, u = 0?

Step 6: Calculate the initial velocity, u

$v e l o c i t y = \frac{d i s p l a c e m e n t}{t i m e}$

$u = \frac{10}{0.5} = 20 m s^{- 1}$

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?

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

Answer:

Step 1: List the known quantities

Initial velocity, u = 12sin(50) m s^–1
Acceleration, a = –9.81 m s^–2
Final velocity, v = 0
Vertical displacement, s = H = ?

Step 2: State the relevant equation of motion

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

Step 3: Rearrange for the displacement, s

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

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

Step 4: Substitute in the values

$s = \frac{{(12 \sin (50))}^{2}}{19.62} = 4.3 m$

Examiner Tip

In the last question, the final velocity can be set to zero because we are only concerned with the object's motion up until it reaches point Q. This is a mathematical trick that simplifies the calculation.

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

You must be confident with resolving velocity (velocity, displacement) into their horizontal and vertical components. Revise this in the Scalars and Vectors topic.

Projectile Motion (CIE AS Physics)

Revision Note

Projectile motion

Projectile motion calculation example

Worked example

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Worked example

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Author: Ashika