1.  In order to solve these problems you need to know:  How to separate a vector into its components.  Quadratic Formula and how to solve it.  Have an understanding of how parabolas work.

2. θ Hypotenuse (h) Adjacent (a) Opposite (o) The length (or in vector terms, magnitude) of the adjacent and opposite sides of this right triangle can be written as the following: This follows from the equations to the left after some algebraic manipulation. This is also the underlying reason why in the video I was able to break the velocity into its component parts.

3. With this concept from trigonometry, let’s see how we can convert the velocity into components. θ V i = 30 m/s V i,x = V i cos( θ) V i,y = V i sin( θ)

4. Whenever we have an equation of the form: We can use what we call the Quadratic Formula to solve for the variable x. The Quadratic Formula says that x equals: Where a, b, and c are the corresponding coefficients in the first equation. The Quadratic Formula is used to determine information from graphs which we call parabolas. The thing to note with parabolas is that there are always two values for x that correspond to a given y. This property comes in handy when solving projectile motion problems.