# A projectile is shot from the ground at an angle of pi/12  and a speed of 3/5 m/s. When the projectile is at its maximum height, what will its distance from the starting point be?

Mar 24, 2017

The distance is $= 0.009 m$

#### Explanation:

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

$u = {u}_{0} \sin \theta = \frac{3}{5} \sin \left(\frac{1}{12} \pi\right)$

$a = - g$

$v = 0$ at the maximum height

We apply the equation

$v = u + a t$

$t = \frac{v - u}{a} = \frac{0 - \frac{3}{5} \sin \left(\frac{1}{12} \pi\right)}{- g}$

$= \frac{3}{5 g} \sin \left(\frac{1}{12} \pi\right) = 0.016 s$

Time to reach the maximum height is $= 0.016 s$

Solving in the horizontal direction $\to +$

distance $d = {u}_{0} t \cos \left(\frac{1}{12} \pi\right) = 0.016 \cdot \frac{3}{5} \cdot \cos \left(\frac{1}{12} \pi\right)$

$= 0.009 m$