# A projectile is shot from the ground at an angle of pi/8  and a speed of 5 /9 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?

Aug 1, 2017

$\text{distance} = 0.0114$ $\text{m}$ $= 1.14$ $\text{cm}$

#### Explanation:

We're asked to find the distance an object is from its launch point when it reaches its maximum height, given its initial velocity.

Vertical:

When the particle is at its maximum height, its instantaneous $y$-velocity is $0$, and we can use the equation

${\left({v}_{y}\right)}^{2} = {\left({v}_{0} \sin {\alpha}_{0}\right)}^{2} - 2 g h$

to find the height $h$

Plugging in known values (${v}_{0} = \frac{5}{9}$ $\text{m/s}$, ${\alpha}_{0} = \frac{\pi}{8}$), we have

0 = ((5/9color(white)(l)"m/s")sin(pi/8))^2 - 2(9.81color(white)(l)"m/s"^2)h

h = ul(0.00230color(white)(l)"m"

The time $t$ when this occurs is given by

${v}_{y} = {v}_{0} \sin {\alpha}_{0} - g t$

$0 = \left(\frac{5}{9} \textcolor{w h i t e}{l} {\text{m/s")sin(pi/8) - (9.81color(white)(l)"m/s}}^{2}\right) t$

$t = 0.0217$ $\text{s}$

Horizontal:

We can find the horizontal distance covered by the equation

$x = {v}_{0} \cos {\alpha}_{0} t$

We found that $t = 0.0217$ $\text{s}$, so we have

$x = \left(\frac{5}{9} \textcolor{w h i t e}{l} \text{m/s")cos(pi/8)(0.0217color(white)(l)"s}\right)$

= ul(0.0111color(white)(l)"m"

Distance:

The distance found via

$r = \sqrt{{x}^{2} + {h}^{2}} = \sqrt{{\left(0.0111 \textcolor{w h i t e}{l} \text{m")^2 + (0.00230color(white)(l)"m}\right)}^{2}}$

= color(red)(ul(0.0114color(white)(l)"m" = color(red)(ul(1.14color(white)(l)"cm"