# A projectile is shot from the ground at an angle of pi/4  and a speed of 6 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?

Apr 17, 2016

$2.6 m$

#### Explanation:

Split up the velocity into its horizontal and vertical components using trigonometry,

${V}_{h} = 6 \frac{m}{s} \cdot \cos \left(\frac{\pi}{4}\right) = 6 \cos 45$
${V}_{v} = 6 \frac{m}{s} \cdot \sin \left(\frac{\pi}{4}\right) = 6 \sin 45$

Now you should work out the halfway point, where upwards velocity becomes $0$ and the projectile reaches its maximum point

$v = u + a t$
$0 = {V}_{v} + \text{gt}$,

where the acceleration of gravity $g = - 9.8$

$0 = 6 \sin 45 - 9.8 t$

$t = \frac{6 \sin 45}{9.8}$
$t = 0.433 s$

Now you can work out the horizontal distance by

$s = v t$
$s = {V}_{h} \cdot 0.433$
$= 6 \cos 45 \cdot 0.433 = 1.84 m$

and the vertical distance by

$s = {V}_{v} \cdot 0.433$
$= 6 \sin 45 \cdot 0.433 = 1.84 m$.

The overall distance is found by Pythagoras' theorem,

$c = \sqrt{{a}^{2} + {b}^{2}}$

$c = \sqrt{{1.84}^{2} + {1.84}^{2}} = 2.6 m$