# A projectile is shot from the ground at an angle of (5 pi)/12  and a speed of 7 /3 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?

Jan 12, 2017

The distance $= 0.14 m$

#### Explanation:

We start by calcating the time to reach the maximum height

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

$u = \frac{7}{3} \sin \left(\left(\frac{5}{12}\right) \pi\right)$

$v = 0$

$a = - g = - 9.8$

We use the equation

$v = u + a t$

$0 = \frac{7}{3} \sin \left(\left(\frac{5}{12}\right) \pi\right) - 9.8 t$

$9.8 t = \frac{7}{3} \sin \left(\left(\frac{5}{12}\right) \pi\right)$

$t = \frac{1}{9.8} \cdot \frac{7}{3} \sin \left(\left(\frac{5}{12}\right) \pi\right) = 0.23 s$

Computing in the horizontal direction ${\rightarrow}^{+}$

$u = \frac{7}{3} \cos \left(\left(\frac{5}{12}\right) \pi\right)$

$d =$ distance from starting point

$d = \frac{7}{3} \cos \left(\left(\frac{5}{12}\right) \pi\right) \cdot 0.23 = 0.14 m$