PlanetPhysics/Projectile Motion

Consider the motion of a particle which is projected in a direction making an angle $$\alpha$$ with the horizon. When we neglect drag, the only force which acts upon the particle is its weight, $$m{\mathbf g}$$ (Fig. 66).

\begin{figure} \includegraphics[scale=.8]{Fig66.eps} \end{figure}

Taking the plane of motion to be the xy-plane, and applying Newton's laws of motion gives us the equations

{\mathbf x-axis} \\

$$ m \frac{d \ddot{x}}{dt} = 0$$ $$ \frac{d \ddot{x}}{dt} = 0 $$ {\mathbf y-axis} \\

$$ m \frac{d \ddot{y}}{dt} = -mg$$ $$ \frac{d \ddot{y}}{dt} = -g $$ where $$\frac{d \ddot{x}}{dt}$$ and $$\frac{d \ddot{y}}{dt}$$ are the components of the acceleration along the x and y axes. Integrating equations (1) and (2) we get

$$ \dot{x} = c_1 $$ $$ \dot{y} = -gt + c_2 $$

Therefore the component of the velocity along the x-axis remains constant, while the component along the y-axis changes uniformly. Let $$v_0$$ be the initial velocity of the projection, then when $$t = 0$$, $$\dot{x}_0 = v_0 \cos \alpha$$ and $$\dot{y}_0 = v_0 \sin \alpha$$. Making these substitutions in the last two equations we obtain

$$ c_1 = v_0 \cos \alpha $$ $$ c_2 = v_0 \sin \alpha $$

Therefore

$$ \dot{x} = v_0 \cos \alpha $$ $$ \dot{y} = v_0 \sin \alpha - gt $$ Then the total velocity at any instant is

$$ v = \sqrt{\dot{x}^2 + \dot{y}^2} $$ $$ v = \sqrt{v_0^2 - 2v_0 g t \sin \alpha + g^2 t^2} $$

and makes an angle $$\theta$$ with the horizon defined by

$$ \tan \theta = \frac{ \dot{y} }{ \dot{x} } = \frac{ v_0 \cos \alpha}{ v_0 \sin \alpha - gt} $$

Integrating equations (3) and (4) we obtain

$$ x = v_0 t \cos \alpha + c_3 $$ $$ y = v_0 t \sin \alpha - \frac{1}{2} g t^2 + c_4 $$

But when $$t = 0$$, $$x = y = 0 $$, therefore $$c_3 = c_4 = 0$$, and consequently

$$ x = v_0 t \cos \alpha $$ $$ y = v_0 t \sin \alpha - \frac{1}{2} g t^2 $$ It is interesting to note that the motions in the two directions are independent. The gravitational acceleration does not affect the constant velocity along the x-axis, while the motion along the y-axis is the same as if the body were dropped vertcally with an initial velocity $$v_0 \sin \alpha $$.

{bf The Path} - The equation of the path may be obtained by eliminating $$t$$ between equations (7) and (8). This gives

$$ y = x \tan \alpha - \frac{g}{2 v_0^2 \cos^2 \alpha} x^2 $$ which is the equation of a parabola.

{bf The Time of Flight} - When the projectile strikes the ground its y-coordinate is zero. Therefore substituting zero for $$y$$ in equation (8) we get for the time of flight

$$ T = \frac{2 v_0 \sin \alpha}{g} $$ {\mathbf The Range} - The range, or the total horizontal distance covered by the projectile, is found by replacing $$t$$ in equation (7) by the value of $$T$$ in equation (10), or by letting $$y = 0$$ in equation (9). By either method we obtain

$$ R = \frac{2 v_0^2 \sin \alpha \cos \alpha}{g} = \frac{v_0^2 \sin 2 \alpha}{g} $$ Note that a basic trigonometric identity was used to simpilfy the above equation.

Since $$v_0$$ and $$g$$ are constants the value of $$R$$ depends upon $$\alpha$$. It is evident from equation (11) that $$R$$ is maximum when $$\sin 2 \alpha = 1$$, or when $$\alpha = \frac{\pi}{4}$$. The maximum range is, therefore,

$$ R_{max} = \frac{v_0^2}{g} $$ In actual practice the angle of elevation which gives the maximum range is smaller on account of the resistance of the air.

{\mathbf The Highest Point} - At the highest point $$\dot{y}=0$$. Therefore substituting this value of $$\dot{y}$$ in equation (4) we obtain $$\frac{v_0 \sin \alpha}{g}$$ or $$\frac{1}{2}T$$ for the time taken to reach the highest point. Subsituting this value of the time in equation (8) we get for the maximum elevation

$$ H = \frac{v_0^2 \sin^2 \alpha}{2 g} $$ {\mathbf The Range for a Sloping Ground} - Let $$\beta$$ be the angle which the ground makes with the horizon. Then the range is the distance $$OP$$, Fig. 67, where $$P$$ is the point where the projectile strikes the sloping ground. The equation of the line $$OP$$ is

\begin {equation} y = x \tan \beta \begin{figure} \includegraphics[scale=.8]{Fig67.eps} \end{figure}

Eliminating $$y$$ between equations (14) and (9) we obtain the x-coordinate of the point,

$$ x_p = \frac{ 2 v_0^2 \cos^2 \alpha \left ( \tan \alpha - \tan \beta \right ) }{g} $$

But $$x_p = R' \cos \beta$$, where $$R' = OP$$.

Therefore

$$ R' = \frac{2 v_0^2 \cos \alpha}{g \cos^2 \beta} \sin \left ( \alpha - \beta \right ) $$

$$ R' = \frac{ v_0^2}{g} \frac{ \sin \left ( 2 \alpha - \beta \right ) - \sin \beta }{ \cos^2 \beta } $$ Thus for a given value of $$\beta$$, $$R'$$ is maximum when $$\sin \left ( 2 \alpha - \beta \right ) = 1$$, that is, when $$\alpha = \frac{\pi}{4} + \frac{\beta}{2}$$.

$$ R_{max}' = \frac{ v_0^2 1 - \sin \beta}{ g \cos^2 \beta} = \frac{v_0^2}{g \left ( 1 + \sin \beta \right ) } $$ When $$\beta = 0$$ equations (15) and (16) reduce to equations (12) and (13), as they should.