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?

1 Answer
Mar 11, 2016

~~5.54s

Explanation:

velocity of projection,u=64ms^-1
angle of projection,alpha=2pi/3
if time of reaching maximum height be t
then it will have zero velocity at the peak.
So0=u*sinalpha- g*t
=>t=u*sinalpha/ g=64*sin(2pi/3)/10=6.4*sqrt3/2=3.2*sqrt3m~~5.54s