Trajectory of a Projectile on a Hill

Written by Jerry Ratzlaff on . Posted in Classical Mechanics

trajectory hill

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

 

Tags: Equations for Velocity