# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 7 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 11, 2017

Distance $= 1.25 m$

#### Explanation:

Solving in the vertical direction ${\uparrow}^{+}$

$u = {u}_{0} \sin \theta = 7 \sin \left(\frac{5}{12} \pi\right)$

$a = - g$

$v = 0$ at the maximum height

We apply the equation

$v = u + a t$

$t = \frac{v - u}{a} = \frac{0 - 7 \sin \left(\frac{5}{12} \pi\right)}{- g}$

$= \frac{7}{g} \sin \left(\frac{5}{12} \pi\right) = 0.69 s$

Time to reach the maximum height is $= 0.69 s$

Solving in the horizontal direction $\to +$

distance $d = {u}_{0} t \cos \left(\frac{5}{12} \pi\right) = 0.69 \cdot 7 \cdot \cos \left(\frac{5}{12} \pi\right)$

$= 1.25 m$