What is projectile motion?

IB physics is a demanding subject, regardless of whether students are studying it at HL or SL. One subject area that students often struggle with in IB physics is projectile motion, or mechanics. However, with reasonable consolidation of IB material, mechanics doesn’t have to be intimidating to IB physics students.

In its most simple terms, projectile motion allows us to see the motion of objects when they are being acted on by the force of gravity. In projectile motion, we have to consider both the horizontal motion and vertical motion of the projectile.

At IB-level, physics students are expected to understand how to manipulate equations, using them to model motion, and respond to multiple-part questions.

Using equations to model motion at IB-level

No forces affect the horizontal motion. However, the vertical motion is affected by gravity and we can assume that no other forces affect it, including air resistance and rotational movement.

This means we can take the vertical acceleration of the particle to be equal to $-g$, where $g=9.8\,ms^{-2}$. This can be modelled by the following equations:

Horizontal velocity : $v_{x}=u\,cos\theta$

Vertical velocity : $v_{y}=u\,sin\theta$

Where $\theta$ is the angle at which the object is projected, $u$ is the initial velocity and $t$ is the time taken.

Next, the position of the projectile needs to be considered. Similar to before, both the horizontal and vertical components of the displacement should be evaluated. Using the formulas:

$x=ut\,cos\theta$ $y=ut\,sin\theta - \frac{1}{2}gt^{2}$

Then, these 4 equations can be combined to find the overall final velocity using $v^{2}=u^{2} - 2as$, an equation that IB students should be familiar with, thanks to their units on velocity and acceleration. Using these equations, IB students should be able to calculate any component of the projectile motion, including the time, velocity and distance, depending on the context of the question.

Answering IB physics questions on mechanics

Most often, students can grasp the basic concepts of mechanics in IB physics, but struggle a bit more when it comes to applying their knowledge to questions. Let’s take a look at an example question and response:

An object is launched horizontal from a height of $20 m$ above the ground with speed $15\,ms^{-1}$. Determine:

a) The time at which it will hit the ground

The launch is horizontal, i.e. $\theta=0^{°}$, and so the formula for the vertical displacement is $y=ut\,sin\theta - \frac{1}{2}gt^{2}$, since $sin\,0 =0$, we can rewrite the displacement as $y=-\frac{1}{2}gt^{2}$. The object will hit the ground when $y=-20m$, so we can substitute this value into the formula.

$y=-\frac{1}{2}gt^{2}$

$-20=-\frac{1}{2}(9.8)t^{2}$

$-20=-4.9t^{2}$

$4.081=t^{2}$

$t=\sqrt{4.081}$

$t=2.02\,seconds$

b) The horizontal distance travelled

The horizontal distance can be found using formulas we mentioned before, $x=ut\,cos\theta$.

So, after substituting in values, the answer is:

$x=15 \times 2.02 \times cos\,0$

$x=30.3\,metres$

c) The speed at which it hits the ground.

Using the formula $v^{2}=u^{2} - 2as$ to get:

$v^{2}=15^{2}-2 \times 9.8 \times -20$

$v^{2}=617$

$v^{2}=\sqrt{617}$

$v=24.8\,ms^{-1}$

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