Projectiles Launched Obliquely

Projectiles launched obliquely means there is both a vertical component, and a horizontal component during the projectile’s launch. This implies that there would be both an initial vertical velocity and an initial horizontal velocity, whereas projectiles launched horizontally only had an initial horizontal velocity.

Calculations

The 2 equations that you’ll use in every projectile motion are very simple and stated below:

A good trick to remember which one is which is by thinking of which direction you plant a sign; vertically. Hence sine is used for vertical velocities, and so cosine has to be used for horizontal velocities.

The same equations apply for scenarios in which projectiles launched obliquely as scenarios in the previous notes page which covered projectiles launched horizontally (read that page if you haven’t already, before reading this page). The only difference in this case is that we do have an initial vertical velocity which is not equal to 0. To solve problems, you’d need to use a combination of different equations depending on the variables you are given. It is always helpful to write down all the variables given in the question, so you can easily figure out which equation(s) you need to use.

Tips for Solving Problems

There are a couple of key things you need to remember when tackling projectile motion problems. They are stated below:

• The maximum height of a projectile is achieved when the final vertical velocity is 0. Applying this to the equation [v(v)]^2 = [u(v)]^2 + 2as(v) will allow you to easily calculate the maximum height, given you know the initial vertical velocity.

• If a projectile lands a certain distance, s(h), on level ground, this then means that s(v) at the end is 0. This information can be applied to s(v) = u(v)t + 1/2 at^2 to find the time it took for the projectile to land, given you know the initial vertical velocity.

• For some problems, you might need to use more than 1 equation to arrive at the final answer. So don’t get dis-heartened if only one equation doesn’t give you the answer. It might end up giving you the value of a variable which you can use in a second equation to get the final answer. For example, you might be asked to find the height of a projectile when it’s a certain horizontal distance away from where it was launched. You are only given the initial velocity, the angle at which it was launched and the horizontal distance (at which point it wants to know what the height of the projectile is). Clearly s(v) = u(v)t+ 1/2 at^2 won’t give you the answer because you don’t know the time, t. So you can use u(h) = s(h)/t (horizontal speed = horizontal distance / time) to first find the time for the projectile to reach the horizontal distance given, and then you can use that time to find the height.

Graphs

Imagine a projectile being launched as shown in the diagram below:

If you were to graph vertical acceleration vs time, vertical displacement vs time, vertical velocity vs time, horizontal velocity vs time and vertical speed vs time, they would look like the graphs below:

Projectile motion worked example | Elucidate Video

In the above video, Dev thoroughly explains how to breakdown and solve a typical projectile motion question. The steps demonstrated in the video can be applied to most projectile motion questions.