# A projectile is shot from the ground at an angle of pi/8  and a speed of 4 /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?

Sep 6, 2017

We're not given the projectile's coefficient of drag, which we'd need to include to take wind resistance into consideration. So, we ignore this.

the vertical component of the projectile's velocity is:

$\sin \left(\frac{\pi}{8}\right) \cdot \frac{4}{9} \frac{m}{s}$

The projectile's vertical velocity, expressed as a function of t, will then be:

${v}_{t} = \frac{4}{9} - 9.81 t$

And at max height, this is zero, so we can solve for the time when this occurs:

$\frac{4}{9} - 9.81 t = 0$
$\frac{4}{9} = 9.81 t$
$t = \frac{4}{9 \cdot 9.81} = 0.0453 \sec$

Distance travelled is the horizontal velocity multiplied by this time value. Horizontal velocity is the initial velocity times the cosine of the launch angle.

$d = 0.0453 \cdot \cos \left(\frac{\pi}{8}\right) \cdot \frac{4}{9}$

$= .00791 m$ - not very far!