Trajectory of a Projectile on a Hill

on . Posted in Classical Mechanics

  

trajectory hill

Trajectory of a Projectile on a Hill formulas

\(  t =  ( 2 \; v_{0h} \;/\; g )  \pm   \sqrt{  ( v_{0h}^2 \;/\; g^2 ) - ( 2 \;h \;/\; g ) } \)

\(  d =   v_{0d} \; t \) 

\( h =   v_{0h} \; t - \frac{1}{2} \; g \; t^2  \) 

Symbol English Metric
\( t \) = time \(sec\) \(s\)
\( g \) = gravitational acceleration \(ft\;/\;sec^2\) \(m\;/\;s^2\)
\( v_0 \) = launch velocity \(ft\;/\;sec\) \(m\;/\;s\)
\( \theta \) = vertical angle \(rad\) \(rad\)

 

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Tags: Velocity Projectile