# If a projectile is shot at an angle of pi/6 and at a velocity of 23 m/s, when will it reach its maximum height??

Jan 6, 2016

$h = 6.747 m$

#### Explanation:

Data:-
Angle$= \theta = \frac{\pi}{6}$
Initial Velocity$=$Muzzle Velocity$= {v}_{0} = 23 \frac{m}{s}$
Acceleration due to gravity$= g = 9.8 \frac{m}{s} ^ 2$
Maximum Height=h=??

Sol:-
We know that:
$h = \frac{{v}_{0}^{2} {\sin}^{2} \theta}{2 g}$
$\implies h = \frac{{23}^{2} \cdot {\sin}^{2} \left(\frac{\pi}{6}\right)}{2 \cdot 9.8} = \frac{529 \cdot 0.25}{19.6} = \frac{132.25}{19.6} = 6.747 m$
$\implies h = 6.747 m$