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 1Trajectory of a Projectile formula

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

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 1Horizontal range of a Projectile formula

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

Where:

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

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

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

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

 

 

trajectory range 1Launch Velocity of a Projectile formula

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

Where:

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

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

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

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

 

 

trajectory max heightMaximum Height of a Projectile formula

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

Where:

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

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

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

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

 

 

trajectory clearing 1Projectile Clearing an Object formula

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

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 1Projectile Launch Angle formula

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

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 timeTime of flight of a Projectile formula

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

Where:

\(\large{ t }\) = time

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

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

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

 

 

trajectory hillTrajectory of a Projectile on a Hill formula

\(\large{  d =   v_{0d} \; t }\)

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

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

Where:

\(\large{ t }\) = time

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

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

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

 

Tags: Equations for Velocity