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.

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