# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 15 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?

Mar 2, 2016

$\text{1) "x=v_i*t*cos alpha " horizontal movement}$
$\text{2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement}$
$\text{3) } x \cong 5 , 734 m$

#### Explanation: ${v}_{i} : 15 \frac{m}{s} \text{ initial velocity}$
$\alpha : \frac{5 \cancel{\pi}}{12} \cdot \frac{180}{\cancel{\pi}} = {75}^{o} \text{ sin75~=0,966 cos75~=0,259}$
${v}_{x} = {v}_{i} \cdot \cos 75 \text{ x component of initial velocity}$
${v}_{y} = {v}_{i} . \sin 75 \text{ y component of initial velocity}$
${t}_{m} = {v}_{i} \cdot \sin \frac{\alpha}{g} \text{elapsed time to reach to max height}$
${h}_{m} = \frac{{v}_{i}^{2} \cdot {\sin}^{2} \alpha}{2} \cdot g \text{ maximum height}$
$\text{1) "x=v_i*t*cos alpha " horizontal movement}$
$\text{2) "y=v_i*t*sin alpha-1/2*g*t^2" vertical movement}$
"3) "t_m=(15*sin75)/(9,81~)=1,477 s
$x = 15 \cdot \cos 75 \cdot 1 , 477$
$x \cong 5 , 734 m$