Trajectory of a Projectile

Written by Jerry Ratzlaff on . Posted in Classical Mechanics

Trajectory of a projectile is the curved path given an initial velocity and is acted on by gravity.  The projectile is acted upon by both vertical and horizontal motion along the trajectory.

 

trajectory angle 1Formulas that use Trajectory of a Projectile

\(\large{  y = d \; tan \; \theta - \frac{ g \; d^2  }{ 2 \; v_0^2 \; cos^2 \; \theta } }\)   

Where:

\(\large{ y }\) = vertical position

\(\large{ \theta }\) = angle of the initial vertical from x-axis

\(\large{ g }\) = gravitational acceleration

\(\large{ d }\) = horizontal position

\(\large{ v_0 }\) = launch velocity

 

trajectory range 1Formulas that use Horizontal range of a Projectile

\(\large{  R =  \frac{ v_0^2 \; sin \; 2 \theta  }{ g } }\)   

Where:

\(\large{ R }\) = horizontal range

\(\large{ g }\) = gravitational acceleration

\(\large{ v_0 }\) = launch velocity

\(\large{ \theta }\) = vertical angle

 

 

trajectory range 1Formulas that use Launch Velocity of a Projectile

\(\large{  v_0 =   \sqrt{  \frac{ R \; g }{ sin\; 2\theta }  }   }\)   

Where:

\(\large{ v_0 }\) = launch velocity

\(\large{ R }\) = horizontal range

\(\large{ g }\) = gravitational acceleration

\(\large{ \theta }\) = vertical angle

 

 

trajectory max heightFormulas that use Maximum Height of a Projectile

\(\large{  h_{max} =  \frac{ v_0 \; sin^2 \; \theta  }{ 2 \; g } }\)   

Where:

\(\large{ h_{max} }\) = maximum height

\(\large{ g }\) = gravitational acceleration

\(\large{ v_0 }\) = launch velocity

\(\large{ \theta }\) = vertical angle

 

 

trajectory clearing 1Formulas that use Projectile Clearing an Object

\(\large{  y_h = \frac{d \; v_{0y} }{v_{0x} } - \frac{1}{2} \; g \; \frac{x^2}{v_{0x}^2}  }\)   

Where:

\(\large{ y_h }\) = vertical position

\(\large{ \theta }\) = angle of the initial vertical from x-axis

\(\large{ g }\) = gravitational acceleration

\(\large{ d }\) = horizontal position

\(\large{ v_0 }\) = launch velocity

 

trajectory angle 1Formulas that use Projectile Launch Angle

\(\large{  \theta = \frac{ 1 }{ 2 } \; sin^{-1} \left( \frac{ g\; d }{ v_0^2 }  \right)   }\)   

Where:

\(\large{ \theta }\) = angle of the initial vertical from x-axis

\(\large{ g }\) = gravitational acceleration

\(\large{ d }\) = horizontal position

\(\large{ v_0 }\) = launch velocity

 

 

trajectory timeFormulas that use Time of flight of a Projectile

\(\large{  t =  \frac{ 2 \; v_0 \; sin \; \theta  }{ g } }\)   

Where:

\(\large{ t }\) = time

\(\large{ g }\) = gravitational acceleration

\(\large{ v_0 }\) = launch velocity

\(\large{ \theta }\) = vertical angle

 

 

trajectory hillFormulas 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