A projectile is shot at a velocity of 1 m/s and an angle of pi/12 . What is the projectile's maximum height?

Mar 25, 2017

The maximum height is $= 0.00358 m$

Explanation:

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

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

$a = - g$

$v = 0$ at the maximum height

We apply the equation

${v}^{2} = {u}^{2} + 2 a h$

$0 = {\sin}^{2} \left(\frac{\pi}{12}\right) - 2 g h$

$h = \frac{1}{2 g} \cdot {\sin}^{2} \left(\frac{\pi}{12}\right)$

$= 0.00358 m$