# If a projectile is shot at an angle of (2pi)/3 and at a velocity of 64 m/s, when will it reach its maximum height?

Mar 11, 2016

$\approx 5.54 s$
velocity of projection,$u = 64 m {s}^{-} 1$
angle of projection,$\alpha = 2 \frac{\pi}{3}$
So$0 = u \cdot \sin \alpha - g \cdot t$
$\implies t = u \cdot \sin \frac{\alpha}{g} = 64 \cdot \sin \frac{2 \frac{\pi}{3}}{10} = 6.4 \cdot \frac{\sqrt{3}}{2} = 3.2 \cdot \sqrt{3} m \approx 5.54 s$