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ASPhysicsCIERevision Notes2. Kinematics2.1 Equations of MotionProjectile Motion

Projectile Motion (CIE AS Physics)

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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 space equals space u t space plus thin space 1 half a t squared

Step 3: Substitute in the values

L space equals 1 half g t squared

L space equals space 1 half space cross times space 9.81 space cross times space open parentheses 0.5 close parentheses squared space equals space 1.2 space straight 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 space equals space u t thin space plus space 1 half a t squared

Step 3: Rearrange for the time, t

s space equals space 1 half g t squared

t space equals space square root of fraction numerator 2 s over denominator g end fraction end root

Step 4: Substitute the values

t space equals space square root of fraction numerator 2 space cross times space 1.25 over denominator 9.81 end fraction end root equals space 0.5 space straight 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 space equals fraction numerator space d i s p l a c e m e n t over denominator t i m e end fraction

u space equals fraction numerator space 10 over denominator 0.5 end fraction space equals space 20 space straight m space straight s to the power of negative 1 end exponent

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 squared space equals space u squared space plus space 2 a s

Step 3: Rearrange for the displacement, s

2 a s space equals space v squared space minus space u squared

s space equals space fraction numerator v squared space minus space u squared space over denominator 2 a end fraction

Step 4: Substitute in the values

s space equals space fraction numerator open parentheses 12 sin open parentheses 50 close parentheses close parentheses squared space over denominator 19.62 end fraction space equals space 4.3 space straight 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.

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    Ashika

    Author: Ashika

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