# Question #ee158

Jan 24, 2018

Starting at 0, increasing the angle up to $\frac{\pi}{4}$ $\left({45}^{\circ}\right)$ will increase the distance the object travels. After this point up to $\frac{\pi}{2}$ $\left({90}^{\circ}\right)$, increasing the angle decreases the distance travelled.

#### Explanation:

This does intuitively make sense. ${45}^{\circ}$ is the angle half-way from 0 and a right angle, so this should maximise both vertical distance and horizontal height.

Nonetheless, we can also prove this using a bit of algebra and calculus. Even if you can't do the second bit, first bit should make sense. The task is:

Consider this diagram.

Prove that the distance $r = - \frac{2 {u}^{2} \sin \theta \cos \theta}{g}$

Still here? Lets go.

Let's say we are firing a projectile with a velocity of $u$ at an angle $\theta$, and we want to find the horizontal range $r$. The particle is accelerating to the ground with accelerating $g$.
Considering vectors, the vertical velocity of the particle is $u \sin \theta$.

Find the time it takes for the particle to return to the ground if we throw it up and that velocity.
$s = 0$ It's returning to the start.
$u = u \sin \theta$
$v = \text{/}$ we don't care what v is
$a = - g$ since we've chosen up to be the positive direction, this means down must be negative.
$t = t$ this is what we want to find.

Using the equation $s = u t + \frac{1}{2} a {t}^{2}$

$0 = u t \sin \theta + \frac{1}{2} g {t}^{2}$
$0 = \frac{1}{2} t \left(g t + 2 u \sin \theta\right)$
$t = \frac{- 2 u \sin \theta}{g}$

Now, consider the horizontal motion

$s = r$ this is what we want to find
$u = u \cos \theta$
$v = \text{/}$
$a = 0$
$t = - \frac{2 u \sin \theta}{g}$

since $a = 0 , s = u t$

$r = - u \cos \theta \cdot \frac{2 u \sin \theta}{g}$
$r = - \frac{2 {u}^{2} \sin \theta \cos \theta}{g}$

Now, as an extra challenge, can you prove that the angle that gives the maximum r is $\frac{\pi}{4} \left({45}^{\circ}\right)$?

We can use some trig identities to help us with this expression, then differentiate to find the maximum angle.

Since $2 \sin x \cos x = \sin 2 x$;

$r = - {u}^{2} / g \sin \left(2 \theta\right)$

Now, differentiate both sides wrt $\theta$. This gives us the gradient function. When this is equal to zero, we will find the maximum value.
Since $\frac{d}{\mathrm{dx}} \left(\sin k x\right) = k \cos k x$;

$\frac{\mathrm{dr}}{d \theta} = - \frac{2 {u}^{2}}{g} \cos \left(2 \theta\right)$

Let $\frac{\mathrm{dr}}{d \theta} = 0$

$- \frac{2 {u}^{2}}{g} \cos \left(2 \theta\right) = 0$

$\cos \left(2 \theta\right) = 0$

We want an angle between 0 and a right angle, so we want a value in the range $0 \le \theta \le \frac{\pi}{2}$

$2 \theta = \frac{\pi}{2} \text{ or } {90}^{\circ}$

$\theta = \frac{\pi}{4} \text{ or } {45}^{\circ}$