Trajectory of a Projectile

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 of a Projectile formula

$$y = d \; tan \; \theta - ( g \; d^2 \;/\; 2 \; v_0^2 \; cos^2 ( \theta ) \;)$$
Symbol English Metric
$$y$$ = vertical position $$ft$$ $$deg$$
$$\theta$$ = angle of the initial vertical from x-axis $$deg$$ $$rad$$
$$g$$ = gravitational acceleration $$ft\;/\;sec^2$$ $$m\;/\;s^2$$
$$d$$ = horizontal position $$ft$$ $$m$$
$$v_0$$ = launch velocity $$ft\;/\;sec$$ $$m\;/\;s$$

horizontal range of a Projectile formula

$$R = v_0^2 \; sin \; 2 (\theta) \;/\; g$$
Symbol English Metric
$$R$$ = horizontal range $$ft$$ $$\ m$$
$$v_0$$ = launch velocity $$ft\;/\;sec$$  $$m\;/\;s$$
$$\theta$$ = vertical angle $$deg$$ $$rad$$
$$g$$ = gravitational acceleration  $$ft\;/\;sec^2$$   $$m\;/\;s^2$$

Launch Velocity of a Projectile formula

$$v_0 = \sqrt{ R \; g \;/\; sin\; 2 \;( \theta ) }$$
Symbol English Metric
$$v_0$$ = launch velocity $$ft\;/\;sec$$ $$m\;/\;s$$
$$R$$ = horizontal range $$ft$$ $$m$$
$$g$$ = gravitational acceleration $$ft\;/\;sec^2$$ $$m\;/\;s^2$$
$$\theta$$ = vertical angle $$deg$$ $$rad$$

Maximum Height of a Projectile formula

$$h_{max} = v_0 \; sin^2 \; (\theta) \;/\; 2 \; g$$
Symbol English Metric
$$h_{max}$$ = maximum height $$ft$$ $$m$$
$$v_0$$ = launch velocity $$ft\;/\;sec$$ $$m\;/\;s$$
$$\theta$$ = vertical angle $$deg$$ $$rad$$
$$g$$ = gravitational acceleration $$ft\;/\;sec^2$$ $$m\;/\;s^2$$

Projectile Clearing an Object formula

$$y_h = ( d \; v_{0y} \;/\; v_{0x} ) - [\; \frac{1}{2} \; g \; ( x^2\;/\;v_{0x}^2 ) \;]$$
Symbol English Metric
$$y_h$$ = vertical position $$ft$$ $$m$$
$$d$$ = horizontal position $$ft$$ $$m$$
$$v_0$$ = launch velocity $$ft\;/\;sec$$ $$m\;/\;s$$
$$g$$ = gravitational acceleration $$ft\;/\;sec^2$$ $$m\;/\;s^2$$
$$\theta$$ = angle of the initial vertical from x-axis $$deg$$ $$rad$$

Projectile Launch Angle formula

$$\theta = \frac{ 1 }{ 2 } \; sin^{-1} \left( g\; d \;/\; v_0^2 \right)$$
Symbol English Metric
$$\theta$$ = angle of the initial vertical from x-axis $$deg$$ $$rad$$
$$g$$ = gravitational acceleration  $$ft\;/\;sec^2$$   $$m\;/\;s^2$$
$$d$$ = horizontal position $$ft$$ $$m$$
$$v_0$$ = launch velocity $$ft\;/\;sec$$  $$m\;/\;s$$

Time of Flight of a Projectile formula

$$t = 2 \; v_0 \; sin (\theta) \;/\; g$$
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 $$deg$$ $$rad$$

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 $$rad$$ $$rad$$

Tags: Velocity Projectile