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Calculating Trajectories

Applied to the case of trajectories close to the Earth's surface,^6.17 the equations governing constant horizontal velocity superimposed upon constant downward acceleration take the form

$$\ddot{x}$$ = 0 (6.4)

$$\dot{x} v_{x_0} \quad \hbox{\rm (constant)}$$ = (6.5)

$$x_0 \; + \; v_{x_0} \, t$$ x = (6.6)

$$\hbox{\rm and} \qquad \qquad \qquad \qquad \qquad \qquad$$ (6.7)

$$\ddot{y}$$ = - g (6.8)

$$\dot{y} v_{y_0} \; - \; g \, t$$ = (6.9)

$$y_0 \; + \; v_{y_0} \, t \; - \; {1\over2} \, g \, t^2$$ y = (6.10)

where

$$\ddot{x} \equiv {d^2 x \over dt^2} \equiv {dv_x \over dt} \equiv \dot{v_x} \equiv a_x ,$$ (6.11)

$$\dot{x} \equiv {dx \over dt} \equiv v_x ,$$ (6.12)

$$\ddot{y} \equiv {d^2 y \over dt^2} \equiv {dv_y \over dt} \equiv \dot{v_y} \equiv a_y$$ (6.13)

$$\hbox{\rm and} \qquad \dot{y} \equiv {dy \over dt} \equiv v_y$$ (6.14)

Hold it! Before you bolt for the door, take a moment to casually read through all these horrible-looking equations. I have made them look long and hirsute on purpose, for two reasons: first, because this way they are in their most general form - i.e. we can be confident that these equations will correctly describe any trajectory problem, but for any actual problem the equations will usually simplify; and second, because this is a sort of practical joke - if you look carefully you will see that the equations are really pretty simple! All those ``$\equiv$'' symbols just mean, ``...another way of putting it, which amounts to exactly the same thing, is....'' That is, they just indicate equivalent notations - or, in the language of linguistics, synonyms. So the latter batch of equations is just reminding you of the convention Physicists use for writing time derivatives: ``dot'' and ``double-dot'' notation. The first batch of equations tells you (in this notation) everything there is to know about the motion: the horizontal [x] motion is not under any acceleration [ $a_x \equiv \ddot{x} = 0$] so the horizontal velocity [ $v_x \equiv \dot{x}$] is constant [ $\dot{x} = v_{x_0}$] and the distance travelled horizontally [x(t)] is just increasing linearly with time t relative to its initial value x₀ - i.e. x = x₀ + v_x0t. The vertical motion differs only in that it includes a constant downward acceleration [ $a_y \equiv \ddot{y} = -g$] which adds a term [-gt] to $\dot{y}$ and another familiar term [ $-{1\over2}gt^2$] to y(t). Note that in every case the whole idea is to get the quantity on the left-hand side [lhs] of the equation equal to an explicit function of t on the right-hand side [rhs].

Let's do a problem to illustrate how these equations work: Suppose we fire a cannon horizontally from the top of a 19.62 m high bluff, imparting an initial velocity v_x0 = 10 m/s to the cannonball. [By the definition of ``horizontal,'' v_y0 = 0.] Where does the ball hit? [We neglect air friction and assume level (horizontal) ground at the bottom of the bluff.]

  

Figure: (a) Sketch of a trajectory problem in which the initial height [ y₀ = 19.62 m] and the initial (horizontal) velocity [ v_x0 = 10 m/s] are given and we want to calculate the horizontal distance [x_f] at which the cannonball hits the ground [y_f = 0]. (b) Corresponding plot of y(x), the trajectory followed by the cannonball.

For simplicity we can take x=0 at the muzzle of the cannon;^6.18 similarly, we (naturally enough) take t=0 to be the instant at which the ball leaves the muzzle of the cannon. Our general equations now ``reduce'' to a more particular set of equations for this specific example:

$$x = v_{x_0} \, t \qquad \hbox{\rm and} \qquad y = y_0 \; - \; {1\over2} \, g \, t^2$$

or, since v_x0 = 10 m/s and y₀ = 19.62 m,

$$x = (10 \hbox{\rm m/s}) \, t \qquad \hbox{\rm and} \qquad y... ... {\rm m}) \; - \; {1\over2} \, (9.81 \hbox{\rm m/s}^2) \, t^2$$

We now have a choice between working out the algebra in the first pair of equations or working out the arithmetic in the second pair. The former is preferable partly because we don't have to ``juggle units'' while we work out the equations (a clumsy process which is usually neglected, leading to equations with numbers but no units, which in turn can lead to considerable confusion) and because solving for x<sub>f</sub> in terms of the two ``parameters'' y₀ and v_x0 [g is also a parameter, although we usually treat it as if it were a constant of Nature] gives an ``answer'' to any such problem with qualitatively similar conditions. Here's the algebra:

$$x = v_{x_0} \, t \qquad \Longrightarrow \qquad t = {x \over v_{x_0}}$$

which can be substituted for t in the second equation, giving

$$y = y_0 \; - \; {1\over2} \, g \, \left[ x \over v_{x_0} \right]^2 .$$

We are interested in the value of x_f at the end of the trajectory -- i.e. when y_f = 0:

$$y_f = 0 = y_0 \; - \; {1\over2} \, g \, \left[ x_f \over v_{x... ...row \qquad y_0 = {1\over2} \, g \, { x_f^2 \over v_{x_0}^2 }$$

$$\Longrightarrow \qquad {2 y_0 \over g} = { x_f^2 \over v_{x_0... ...Longrightarrow \qquad x_f = \sqrt{2 y_0 \, v_{x_0}^2 \over g}.$$

Now we ``plug in'' y₀ = 19.62 m, v_x0 = 10 m/s and g = 9.81 m/s², giving

$$x_f = \sqrt{2 \times 19.62{\rm m} \times [10 {\rm m/s}]^2 \... ...ver 9.81 {\rm m/s}^2 } = \sqrt{400 {\rm m}^2 } = 20 {\rm m}.$$

And that's the answer: x_f = 20 m. Simple, huh?