# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 3/5 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?

Jul 29, 2017

$\text{distance} = 0.0309$ $\text{m}$

#### Explanation:

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

To do this, we can use the equations

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

and

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

to find the vertical and horizontal positions, respectively, when the particle is at its maximum height.

We realize that the instantaneous s$y$-velocity ${v}_{y}$ is $0$ at its maximum height, so we have

$0 = \left(\frac{3}{5} \textcolor{w h i t e}{l} {\text{m/s")sin((5pi)/12) - 2(9.81color(white)(l)"m/s}}^{2}\right) \left(\Delta y\right)$

Deltay = color(red)(0.0295 color(red)("m"

The time it takes for this to happen is given by

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

$0 = \left(\frac{3}{5} \textcolor{w h i t e}{l} {\text{m/s")sin((5pi)/12) - (9.81color(white)(l)"m/s}}^{2}\right) t$

$t = 0.0591$ $\text{s}$

We can now use the second equation to determine the horizontal distance $\Delta x$:

$\Delta x = {v}_{0} \cos {\alpha}_{0} t = \left(\frac{3}{5} \textcolor{w h i t e}{l} \text{m/s")cos((5pi)/12)(0.0591color(white)(l)"s}\right)$

Deltax = color(green)(0.00917 color(green)("m"

The distance from the launch point is thus

$r = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}} = \sqrt{{\left(\textcolor{g r e e n}{0.00917 \textcolor{w h i t e}{l} \text{m"))^2 + (color(red)(0.0295color(white)(l)"m}}\right)}^{2}}$

= color(blue)(ul(0.0309color(white)(l)"m"