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.

Formulas 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

Formulas 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

Formulas 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

Formulas 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

Formulas 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

Formulas 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

Formulas 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

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