# Trajectory of a Projectile on a Hill

Written by Jerry Ratzlaff on . Posted in Classical Mechanics

## Formulas that use Trajectory of a Projectile on a Hill

 $$\large{ d = v_{0d} \; t }$$ $$\large{ h = v_{0h} \; t - \frac{1}{2} \; g \; t^2 }$$ $$\large{ t = \frac{ 2 \; v_{0h} }{ g } \pm \sqrt{ \frac{ v_{0h}^2 }{ g^2 } - \frac{ 2 \;h }{ g } } }$$

### Where:

$$\large{ t }$$ = time

$$\large{ g }$$ = gravitational acceleration

$$\large{ v_0 }$$ = launch velocity

$$\large{ \theta }$$ = vertical angle