Projectile Motion Concepts

When an object is launched through the air at an angle, its motion becomes two-dimensional – it moves both horizontally and vertically at the same time. We usually call this type of movement projectile motion.

To understand and analyze this motion, we will often need to separate the initial velocity into horizontal and vertical components, transforming a complex 2D problem into two simpler 1D problems.

Breaking Down a Velocity Vector

Consider a projectile launched with initial velocity $\vec{v_0}$ at an angle $\theta$ above the horizontal. Using trigonometry, we can express this velocity as two perpendicular components:

the horizontal component:

$v_{0x} = v_0 \cos(\theta)$,

and the vertical component:

$v_{0y} = v_0 \sin(\theta)$.

These components represent the projectile's initial speed in each direction independently.

Finding Magnitude and Angle from Components

Sometimes we're given the components $v_{0x}$ and $v_{0y}$ and need to find the magnitude and direction of the velocity vector. We can work backwards using the Pythagorean theorem and trigonometry.

Finding the magnitude: Since the horizontal and vertical components form a right triangle with the velocity vector as the hypotenuse, we use:

$v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$

Finding the angle: The angle $\theta$ that the velocity makes with the horizontal can be found using the tangent function:

$\tan(\theta) = \dfrac{v_{0y}}{v_{0x}}$

Therefore:

$\theta = \tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)$

Note: One must be careful if the projectile is launched with non-positive $x$ and $y$ components (for instance launched at a downward angle). In this case, it may be safer to calculate $\tan(\theta) = \dfrac{|v_{0y}|}{|v_{0x}|}$ and use reason to understand where the angle you just calculated lies on the graph, for example: