Projectile Motion (Cambridge (CIE) AS Physics): Revision Note

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.

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

$s=ut+\hspace{0.17em}\frac{1}{2}a{t}^2$

Step 3: Substitute in the values

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

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

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?

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

$s=ut\hspace{0.17em}+\frac{1}{2}a{t}^2$

Step 3: Rearrange for the time, t

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

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

Step 4: Substitute the values

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

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 = ?

Step 6: Calculate the initial velocity, u

$\text{velocity}=\frac{\text{displacement}}{\text{time}}$

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

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?

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

$v^2=u^2+2as$

Step 3: Rearrange for the displacement, s

$2as=v^2-u^2$

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

Step 4: Substitute in the values

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

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 acceleration is −g throughout when upwards is positive

• 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.