A projectile is shot at a velocity of  15 m/s and an angle of pi/4 . What is the projectile's peak height?

Jan 10, 2016

$H e i g h t = 5.7375 m$

Explanation:

Data:-
Muzzle Velocity$= {v}_{0} = 15 \frac{m}{s}$

Angle$= \theta \frac{\pi}{4} =$

Acceleration due to gravity$= g = 9.8 \frac{m}{s} ^ 2$

Height=H=??

Sol:-

We know that:-

$H = \frac{{v}_{0}^{2} {\sin}^{2} \theta}{2 g}$

$\implies H = \frac{{15}^{2} {\sin}^{2} \left(\frac{\pi}{4}\right)}{2 \cdot 9.8} = \frac{\left(225\right) {\left(0.707\right)}^{2}}{19.6} = \frac{225 \cdot 0.4998}{19.6} = \frac{112.455}{19.6} = 5.7375 m$

$\implies H = 5.7375 m$