# Question #f551a

Apr 11, 2017

The horizontal range is $= \frac{4 {u}^{2}}{5 g}$

#### Explanation:

Resolving in the vertical direction ${\uparrow}^{+}$

initial velocity is ${u}_{y} = u \sin \theta$

Acceleration is $a = - g$

At the maximum height, $v = 0$

We apply the equation of motion

${v}^{2} = {u}^{2} + 2 a s$

to calculate the greatest height

$0 = {u}^{2} {\sin}^{2} \theta - 2 \cdot g \cdot h$

$h = \frac{{u}^{2} {\sin}^{2} \theta}{2 g}$

We apply the equation of motion

$v = u + a t$

to calculate the time to reach the greatest height

$0 = u \sin \theta - g \cdot t$

$t = \frac{u}{g} \cdot \sin \theta$

Resolving in the horizontal direction ${\rightarrow}^{+}$

We apply the equation of motion

$s = {u}_{x} \cdot t$

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

$= 2 {u}^{2} / g \cos \theta \sin \theta$

It is given that

$s = 2 h$

Therefore,

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

$\cos \theta \sin \theta = {\sin}^{2} \frac{\theta}{2}$

$\tan \theta = 2$

$\theta = {63.4}^{\circ}$

So,

The horizontal range is

$s = 2 {u}^{2} / g \cos \theta \sin \theta$

$= 2 {u}^{2} / g \cos \left({63.4}^{\circ}\right) \sin \left({63.4}^{\circ}\right)$

$= 2 {u}^{2} / g \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}}$

$s = \frac{4 {u}^{2}}{5 g}$