PlanetPhysics/Projectile Motion

Consider the [[../CosmologicalConstant2/|motion]] of a [[../Particle/|particle]] which is projected in a direction making an angle $\alpha$ with the [[../GenericityInOpenSystems/|horizon]]. When we neglect drag, the only force which acts upon the particle is its weight, $m𝐠$ (Fig. 66).

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

Taking the plane of motion to be the xy-plane, and applying [[../NewtonsLaws/|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/|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/|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

$v=\sqrt{{\dot{x}}^2+{\dot{y}}^2}$ $v=\sqrt{v_0^2-2v_{0}gt\sin \alpha +g^2{t}^2}$

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

$y=x\tan \alpha -\frac{g}{2v_0^2{\cos }^2\alpha }{x}^2$ 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

$T=\frac{2v_0\sin \alpha }{g}$ {\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

$R=\frac{2v_0^2\sin \alpha \cos \alpha }{g}=\frac{v_0^2\sin 2\alpha }{g}$ Note that a basic trigonometric [[../Cod/|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 [[../Conduction/|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

$H=\frac{v_0^2{\sin }^2\alpha }{2g}$ {\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 </math> \begin{figure} \includegraphics[scale=.8]{Fig67.eps} \end{figure}

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

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

But $x_p=R^{\prime }\cos \beta$, where $R^{\prime }=OP$.

Therefore

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

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

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