A projectile is shot from the ground at an angle of pi/12  and a speed of 7 /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 27, 2017

The distance is $= 0.005 m$

Explanation:

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

${v}_{y} = {u}_{0} \sin \theta = \frac{7}{15} \sin \left(\frac{1}{12} \pi\right) m {s}^{-} 1$

$a = - g = - 9.8 m {s}^{-} 2$

$v = 0$ at the maximum height

We apply the equation

$v = u + a t$

$0 = \frac{7}{15} \sin \left(\frac{1}{12} \pi\right) - g \cdot t$

$t = \frac{\frac{7}{15} \sin \left(\frac{1}{12} \pi\right)}{g} = 0.012 s$

This is the time to reach the maximum height

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

${v}_{x} = {u}_{0} \cos \theta = \frac{7}{15} \cdot \cos \left(\frac{1}{12} \pi\right) = 0.451 m {s}^{-} 1$

Distance $= {v}_{x} \cdot t = 0.012 \cdot 0.451 = 0.005 m$