Question #ddd99

Jun 14, 2015

At an angle of $\frac{\pi}{4}$ radians, or ${45}^{\circ}$.

Explanation:

When a projectile is launched at an angle, let's call it $\theta$, with the horizontal, its trajectory will have a vertical component and a horizontal component.

Let's assume that the projectile is launched with an initial velocity of ${v}_{0}$. The initial velocity will have a vertical component and a horizontal component

$\left\{\begin{matrix}{v}_{0 x} = {v}_{0} \cdot \cos \theta \\ {v}_{0 y} = {v}_{0} \cdot \sin \theta\end{matrix}\right.$

Horizontally, the projectile is not acted upon by any force. This means that the vertical component of its initial velocity will remain unchanged during the motion.

This means that you can write

$R = {v}_{0 x} \cdot t$ $\text{ } \textcolor{b l u e}{\left(1\right)}$, where

$R$ - the range of the projectile;
$t$ - the total time of flight.

Vertically, things are not that simple. The projectile will be acted upon by the force of gravity, which means that the vertical component of its initial velocity will be affected by the gravitational accceleration, $g$.

Since the projectile starts at ground level and ends up on ground level, its horizontal displacement will be equal to zero. THis means that you can write

${\underbrace{{d}_{y}}}_{\textcolor{g r e e n}{\text{=0}}} = {v}_{\textrm{0 y}} \cdot t - \frac{1}{2} \cdot g \cdot {t}^{2}$ $\text{ } \textcolor{b l u e}{\left(2\right)}$, where

$t$ - the total time of flight.

Solve this equation for $t$ to get

$- \frac{1}{2} \cdot {t}^{2} + {v}_{0 y} \cdot t = 0 \iff t \cdot \left(- \frac{1}{2} g \cdot t + {v}_{0 y}\right) = 0$

$\left\{\begin{matrix}\cancel{{t}_{1} = \text{0 s}} \\ {t}_{2} = \frac{2 \cdot {v}_{0 y}}{g}\end{matrix}\right.$

The only valid solution is $t = \frac{2 \cdot {v}_{0 y}}{g}$. Use this value in equation $\textcolor{b l u e}{\left(1\right)}$ to get

$R = {v}_{0 x} \cdot \frac{2 \cdot {v}_{0 y}}{g}$

Replace ${v}_{0 x}$ and ${v}_{0 y}$ with their respective values to get

$R = \frac{2}{g} \cdot {v}_{0} \cdot \cos \left(\theta\right) \cdot {v}_{0} \cdot \sin \left(\theta\right) = \frac{2 {v}_{0}^{2}}{g} \cdot \sin \left(\theta\right) \cos \left(\theta\right)$

To get the maxium range, you need the maximum value of

$\sin \left(\theta\right) \cos \left(\theta\right) = \text{max}$

The easiest way to figure this out is to use the following trigonometric identity

$\sin \left(\theta\right) \cos \left(\theta\right) = \frac{\sin \left(\theta + \theta\right) + \sin \left(\theta - \theta\right)}{2}$

$\sin \left(\theta\right) \cos \left(\theta\right) = \frac{\sin \left(2 \theta\right) + {\overbrace{\sin \left(0\right)}}^{\textcolor{g r e e n}{\text{=0}}}}{2} = \sin \frac{2 \theta}{2}$

Since the maximum value of the sinus function is 1, you get

$\sin \left(2 \theta\right) = 1$

Take the arsin from both sides of the equation to get

$\arcsin \left(\sin \left(2 \theta\right)\right) = {\underbrace{\arcsin \left(1\right)}}_{\textcolor{g r e e n}{\text{=} \frac{\pi}{2}}} \iff 2 \theta = \frac{\pi}{2}$

Therefore,

$\theta = \textcolor{g r e e n}{\frac{\pi}{4}}$, or $\theta = \textcolor{g r e e n}{{45}^{\circ}}$