A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 2 m/s. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Jun 23, 2018

The distance is $= 0.216 m$

Explanation:

The initial speed is $u = 2 m {s}^{-} 1$

The angle is $\theta = \frac{5}{12} \pi r a d$

The equation of the trajectory of the projectile is

$y = x \tan \theta - \frac{g {x}^{2}}{2 {u}^{2} {\cos}^{2} \theta}$..............$\left(1\right)$

At the maximum height

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\implies$, $\frac{\mathrm{dy}}{\mathrm{dx}} = \tan \theta - \frac{2 g x}{2 {u}^{2} \cos \theta}$

$= \tan \theta - \frac{g x}{{u}^{2} {\cos}^{2} \theta}$

Therefore,

$\tan \theta - \frac{g x}{{u}^{2} {\cos}^{2} \theta} = 0$

$x = \frac{{u}^{2} \tan \theta {\cos}^{2} \theta}{g} = {u}^{2} \frac{\sin \theta \cos \theta}{g}$

and

Plugging this value of $x$ in $\left(1\right)$

$y = {u}^{2} \frac{\sin \theta \cos \theta}{g} \cdot \tan \theta - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \theta}\right) \cdot {\left({u}^{2} \frac{\sin \theta \cos \theta}{g}\right)}^{2}$

$= \frac{{u}^{2} {\sin}^{2} \theta}{g} - \frac{1}{2} \frac{{u}^{2} {\sin}^{2} \theta}{g}$

$= \frac{1}{2} \frac{{u}^{2} {\sin}^{2} \theta}{g}$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Therefore,

$x = {2}^{2} \cdot \sin \left(\frac{5}{12} \pi\right) \cos \frac{\frac{5}{12} \pi}{9.8} = 0.102 m$

$y = \frac{{2}^{2} {\sin}^{2} \left(\frac{5}{12} \pi\right)}{2 \cdot 9.8} = 0.190$

The distance from the starting point to the greatest height is

$d = \sqrt{{x}^{2} + {y}^{2}} = 0.216 m$