# Mechanics – Chapter 3: Projectile Motion

There are some interesting derivations one can make for projectile motion. Consider the following:

1. Projectile reaches its peak when $v_{y} = 0$ so the time at the peak is:
2. $v_{y} = v_{0y} - gt \Rightarrow t_{peak} = \frac{v_{0y} - v_{y}}{g}$
3. At $v_{y} = 0 \Rightarrow t_{peak} = \frac{v_{0y}}{g}$
4. Since $v_{0y} = v_{0}sin(\theta) \Rightarrow t_{peak} = \frac{v_{0}sin(\theta)}{g}$
5. Since there is no air resistance, the motion of the projectile is symmetric, hence:
6. Total time of flight $t_{flight} = 2t_{peak} = \frac{2v_{0}sin(\theta)}{g}$
7. Since $x = v_{0x}t = v_{0}cos(\theta)t$,
8. The total horizontal displacement should be $x_{max} = v_{0}cos(\theta)t_{flight} \Rightarrow x_{max} = \frac{2v_{0}^2sin(\theta)cos(\theta)}{g}$
9. Since the maximum vertical displacement is reached at $t_{peak}$, $y_{max} = v_{0y}t_{peak} - \frac{1}{2}gt_{peak}^2$
10. $y_{max} = \frac{v_{0}^2sin^2(\theta)}{2g}$

To sum up

• $x_{max} = \frac{2v_{0}^2sin(\theta)cos(\theta)}{g}$
• $y_{max} = \frac{v_{0}^2sin^2(\theta)}{2g}$
• $t_{peak} = \frac{v_{0}sin(\theta)}{g}$
• $t_{flight} = 2t_{peak} = \frac{2v_{0}sin(\theta)}{g}$

But trust me, it’s worth your while to understand how these are derived instead of just memorizing them.