### Instruction:
Next, we treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$ (Figure 2-7). Calculate the projectile's range.

### Response:

Next, we treat projectile motion in two dimensions, first without considering air resistance. Let the muzzle velocity of the projectile be $v_0$ and the angle of elevation be $\theta$ (Figure 2-7). Calculate the projectile's displacement, velocity, and range.
Solution. Using $\mathbf{F}=m \mathrm{~g}$, the force components become
$x$-direction
$$
0=m \ddot{x}
$$
y-direction
$-m g=m \ddot{y}$
$(2.31 b)$
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2 / NEWTONIAN MECHANICS-SINGLE PARTICLE
FIGURE 2-7 Example 2.6.
Neglect the height of the gun, and assume $x=y=0$ at $t=0$. Then
$$
\begin{aligned}
& \ddot{x}=0 \\
& \dot{x}=v_0 \cos \theta \\
& x=v_0 t \cos \theta \\
& y=-\frac{-g t^2}{2}+v_0 t \sin \theta \\
&
\end{aligned}
$$
and
$$
\begin{aligned}
& \ddot{y}=-g \\
& \dot{y}=-g t+v_0 \sin \theta \\
& y=\frac{-g t^2}{2}+v_0 t \sin \theta
\end{aligned}
$$

We can find the range by determining the value of $x$ when the projectile falls back to ground, that is, when $y=0$.
$$
y=t\left(\frac{-g t}{2}+v_0 \sin \theta\right)=0
$$
One value of $y=0$ occurs for $t=0$ and the other one for $t=T$.
$$
\begin{aligned}
\frac{-g T}{2}+v_0 \sin \theta & =0 \\
T & =\frac{2 v_0 \sin \theta}{g}
\end{aligned}
$$
2.4 THE EQUATION OF MOTION FOR A PARTICLE
65
The range $R$ is found from
$$
\begin{aligned}
x(t=T) & =\text { range }=\frac{2 v_0^2}{g} \sin \theta \cos \theta \\
R & =\text { range }=\frac{v_0^2}{g} \sin 2 \theta
\end{aligned}
$$
Notice that the maximum range occurs for $\theta=45^{\circ}$.
Let us use some actual numbers in these calculations. The Germans used a long-range gun named Big Bertha in World War I to bombard Paris. Its muzzle velocity was $1,450 \mathrm{~m} / \mathrm{s}$. Find its predicted range, maximum projectile height, and projectile time of flight if $\theta=55^{\circ}$. We have $v_0=1450 \mathrm{~m} / \mathrm{s}$ and $\theta=55^{\circ}$, so the range (from Equation 2.39) becomes
$$
R=\frac{(1450 \mathrm{~m} / \mathrm{s})^2}{9.8 \mathrm{~m} / \mathrm{s}^2}\left[\sin \left(110^{\circ}\right)\right]=202 \mathrm{~km}
$$
Big Bertha's actual range was $120 \mathrm{~km}$. The difference is a result of the real effect of air resistance.

To find the maximum predicted height, we need to calculated $y$ for the time $T / 2$ where $T$ is the projectile time of flight:
$$
\begin{aligned}
T & =\frac{(2)(1450 \mathrm{~m} / \mathrm{s})\left(\sin 55^{\circ}\right)}{9.8 \mathrm{~m} / \mathrm{s}^2}=242 \mathrm{~s} \\
y_{\max }\left(t=\frac{T}{2}\right) & =\frac{-g T^2}{8}+\frac{v_0 T}{2} \sin \theta \\
& =\frac{-(9.8 \mathrm{~m} / \mathrm{s})(242 \mathrm{~s})^2}{8}+\frac{(1450 \mathrm{~m} / \mathrm{s})(242 \mathrm{~s}) \sin \left(55^{\circ}\right)}{2} \\
& =72 \mathrm{~km}
\end{aligned}
$$
