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Projectiles

 

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A projectile is any object propelled through space by the exertion of a force, which ceases after launch. In a general sense, even a football or baseball may be considered a projectile.

 

Suppose that a projectile is launched upward from ground level, with speed v0, making an angle $\theta$ with the horizontal. Neglecting the effect of air resistance, what is the subsequent trajectory of the projectile?

Our first task is to set up a suitable Cartesian coordinate system. A convenient system is illustrated in Fig. 1. The $z$-axis points vertically upwards (this is a standard convention), whereas the $x$-axis points along the projectile's initial direction of horizontal motion. Furthermore, the origin of our coordinate system corresponds to the launch point. Thus, $z=0$ corresponds to ground level.

Neglecting air resistance, the projectile is subject to a constant acceleration $g=9.81 {\rm m} {\rm s}^{-1}$, due to gravity, which is directed vertically downwards. Thus, the projectile's vector acceleration is written

\begin{displaymath} {\bf a} = (0,0,-g). \end{displaymath} (1)

Here, the minus sign indicates that the acceleration is in the minus $z$-direction (i.e., downwards), as opposed to the plus $z$-direction (i.e., upwards).

Figure 1: Coordinates for the projectile problem
\begin{figure} \epsfysize =2in \centerline{\epsffile{traj.eps}} \end{figure}

What is the initial vector velocity ${\bf v}_0$ with which the projectile is launched into the air at (say) $t=0$? As illustrated in Fig. 1, given that the magnitude of this velocity is $v_0$, its horizontal component is directed along the $x$-axis, and its direction subtends an angle $\theta$ with this axis, the components of ${\bf v}_0$ take the form

\begin{displaymath} {\bf v}_0 = (v_0 \cos\theta, 0, v_0 \sin\theta). \end{displaymath} (2)

Note that ${\bf v}_0$ has zero component along the $y$-axis, which points into the paper in Fig. 1.

Since the projectile moves with constant acceleration, its vector displacement ${\bf s}=(x,y,z)$ from its launch point satisfies

\begin{displaymath} {\bf s} = {\bf v}_0 t + \frac{1}{2} {\bf a} t^2. \end{displaymath} (3)

Making use of Eqs. (1) and (2), the $x$-, $y$-, and $z$-components of the above equation are written
$\displaystyle x$ $\textstyle =$ $\displaystyle v_0 \cos\theta t,$ (4)
$\displaystyle y$ $\textstyle =$ $\displaystyle 0,$ (5)
$\displaystyle z$ $\textstyle =$ $\displaystyle v_0 \sin\theta t - \frac{1}{2} g t^2,$ (6)

respectively. Note that the projectile moves with constant velocity, $v_x =dx/dt = v_0 \cos\theta$, in the $x$-direction (i.e., horizontally). This is hardly surprising, since there is zero component of the projectile's acceleration along the $x$-axis. Note, further, that since there is zero component of the projectile's acceleration along the $y$-axis, and the projectile's initial velocity also has zero component along this axis, the projectile never moves in the $y$-direction. In other words, the projectile's trajectory is 2-dimensional, lying entirely within the $x$-$z$ plane. Note, finally, that the projectile's vertical motion is entirely decoupled from its horizontal motion. In other words, the projectile's vertical motion is identical to that of a second projectile launched vertically upwards, at $t=0$, with the initial velocity $v_0 \sin\theta$ (i.e., the initial vertical velocity component of the first projectile)--both projectiles will reach the same maximum altitude at the same time, and will subsequently strike the ground simultaneously.

Equations (4) and (6) can be rearranged to give

\begin{displaymath} z =x \tan\theta -\frac{1}{2}\frac{g x^2}{v_0^{ 2}} \sec^2\theta. \end{displaymath} (7)

As was first pointed out by Galileo, and is illustrated in Fig. 2, this is the equation of a parabola. The horizontal range $R$ of the projectile corresponds to its $x$-coordinate when it strikes the ground (i.e., when $z=0$). It follows from the above expression (neglecting the trivial result $x=0$) that
\begin{displaymath} R = \frac{2 v_0^{ 2}}{g} \sin\theta \cos\theta = \frac{v_0^{ 2}}{g} \sin 2\theta. \end{displaymath} (8)

Note that the range attains its maximum value,
\begin{displaymath} R_{\rm max} = \frac{v_0^{ 2}}{g}, \end{displaymath} (9)

when $\theta=45^\circ$. In other words, neglecting air resistance, a projectile travels furthest when it is launched into the air at $45^\circ$ to the horizontal.

The maximum altitude $h$ of the projectile is attained when $v_z=dz/dt=0$ (i.e., when the projectile has just stopped rising and is about to start falling). It follows from Eq. (6) that the maximum altitude occurs at time $t_0=v_0 \sin\theta/g$. Hence,

\begin{displaymath} h = z(t_0)= \frac{v_0^{ 2}}{2 g} \sin^2\theta. \end{displaymath} (10)

Obviously, the largest value of $h$,
\begin{displaymath} h_{\rm max} = \frac{v_0^{ 2}}{2 g}, \end{displaymath} (11)

is obtained when the projectile is launched vertically upwards (i.e., $\theta =90^\circ$).

Figure 2: The parabolic trajectory of a projectile
\begin{figure} \epsfysize =2in \centerline{\epsffile{para.eps}} \end{figure}

 

 

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