# A projectile is shot from the ground at an angle of pi/12  and a speed of 9 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Feb 4, 2016

$x = 2 , 06422018349$$m e t e r s$

#### Explanation:

$\theta = \frac{\cancel{\pi}}{12} \cdot \frac{180}{\cancel{\pi}} = {15}^{o}$
$\text{ t is the elapsed time from the starting to maximum point}$
$t = {v}_{i} \cdot \sin \frac{15}{g}$
$t = \frac{9 \cdot 0 , 258819045103}{9 , 81}$
$t = 0 , 237448665232$$s$
$x = {v}_{i x} \cdot t$
${v}_{i x} = {v}_{i} \cdot \cos 15$
${v}_{i x} = 9 \cdot 0 , 965925826289$
${v}_{i x} = 8 , 6933324366$$\frac{m}{s}$
$x = 8 , 6933324366 \cdot 0 , 237448665232$
$x = 2 , 06422018349$$m e t e r s$