# Question #cbeb6

Sep 22, 2017

$x \left(\alpha\right) = {\left(10.4\right)}^{2} / g \setminus \sin 2 \alpha$

Maximum horizontal distance is achieved when $\alpha = \frac{\pi}{4} \setminus \left({45}^{o}\right)$.

#### Explanation:

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

$\left.\begin{matrix}v = u + a t & \text{ where " & s="displacement "(m) \\ s=ut+1/2at^2 & \null & u="initial speed "(ms^-1) \\ s=1/2(u+v)t & \null & v="final speed "(ms^-1) \\ v^2=u^2+2as & \null & a="acceleration "(ms^-2) \\ s=vt-1/2at^2 & \null & t="time } \left(s\right)\end{matrix}\right.$

Horizontal Motion

The projectile will move under constant speed (NB we can still use "suvat" equation with a=0).

The projectile will travel a distance $x$ in time $T$, we must resolve the initial speed in the horizontal direction:

$\left\{\begin{matrix}s = & x & m \\ u = & 10.4 \cos \alpha & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & 0 & m {s}^{-} 2 \\ t = & T & s\end{matrix}\right.$

So applying $s = u t + \frac{1}{2} a {t}^{2}$ we get

$x = 10.4 \cos \alpha T$ ..... [A]

Vertical Motion

The projectile travels under constant acceleration due to gravity. Its total displacement will be zero at the point where the projectile reaches ground level, and travels its maximum distance, and we consider the same time interval, $T$. Considering upwards as positive, and again resolving the initial speed (vertically this time):

$\left\{\begin{matrix}s = & 0 & m \\ u = & 10.4 \sin \alpha & m {s}^{-} 1 \\ v = & \text{Not Required} & m {s}^{-} 1 \\ a = & - g & m {s}^{-} 2 \\ t = & T & s\end{matrix}\right.$

Applying $s = u t + \frac{1}{2} a {t}^{2}$ we have:

$0 = 10.4 \sin \alpha T + \frac{1}{2} \left(- g\right) {T}^{2}$
$\therefore g {T}^{2} - 20.8 \sin \alpha T = 0$
$\therefore T \left(g T - 20.8 \sin \alpha\right) = 0$

Leading to two possible solutions for $T$;

$T = 0$ (Where the projectile leaves the ground)
$T = \frac{20.8 \sin \alpha}{g}$ (Where the projectile returns to the ground

Horizontal Distance Travelled

Using this value of $T$ in equation [A] we see that:

$x \left(\alpha\right) = 10.4 \left(\cos \alpha\right) \cdot \frac{20.8 \sin \alpha}{g}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = 10.4 \left(\cos \alpha\right) \cdot \left(2 \cdot 10.4\right) \frac{\sin \alpha}{g}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{{\left(10.4\right)}^{2} \left(2 \sin \alpha \cos \alpha\right)}{g}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {\left(10.4\right)}^{2} / g \setminus \sin 2 \alpha$ ..... [B]

So we have an expression giving the horizontal distance travelled as a function of the launch angle, $\alpha$:

$x \left(\alpha\right) = {\left(10.4\right)}^{2} / g \setminus \sin 2 \alpha$ ..... [B]

Consistent with the given solution.

We get a maximum distance at a critical point of [B], so let us find any critical points. It is vital to remember that when dealing with calculus of trigonometric functions that we must work in radians. And for this problem we require, $0 \le \alpha < \frac{\pi}{2}$.

Differentiating [B] wrt $\alpha$ we get:

$\frac{\mathrm{dx}}{d \alpha} = {\left(10.4\right)}^{2} / g \setminus \left(2 \cos 2 \alpha\right)$

At a critical point, this derivative is zero, giving:

$\frac{\mathrm{dx}}{d \alpha} = 0 \implies \cos 2 \alpha = 0$
$\therefore 2 \alpha = \frac{\pi}{2} \setminus \setminus \setminus \setminus$ for $0 \le 2 \alpha \le \pi$
$\therefore \alpha = \frac{\pi}{4} \setminus \setminus \setminus \setminus \setminus \setminus$ for $0 \le \alpha \le \frac{\pi}{2}$

Which confirms that the maximum horizontal distance is achieved when $\alpha = \frac{\pi}{4} \setminus \left({45}^{o}\right)$.