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

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

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

$x=v_{0}t\cos \alpha +c_{3}$ $y=v_{0}t\sin \alpha -{\frac {1}{2}}gt^{2}+c_{4}$

But when $t=0$, $x=y=0$, therefore $c_{3}=c_{4}=0$, and consequently

$x=v_{0}t\cos \alpha$ $y=v_{0}t\sin \alpha -{\frac {1}{2}}gt^{2}$ 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 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

$H={\frac {v_{0}^{2}\sin ^{2}\alpha }{2g}}$ {\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,

$x_{p}={\frac {2v_{0}^{2}\cos ^{2}\alpha \left(\tan \alpha -\tan \beta \right)}{g}}$

But $x_{p}=R'\cos \beta$, where $R'=OP$.

Therefore

$R'={\frac {2v_{0}^{2}\cos \alpha }{g\cos ^{2}\beta }}\sin \left(\alpha -\beta \right)$

$R'={\frac {v_{0}^{2}}{g}}{\frac {\sin \left(2\alpha -\beta \right)-\sin \beta }{\cos ^{2}\beta }}$ 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.