A machine gun fires a bullet with an initial velocity of 200m/s at an angle of 60° to the horizontal. If g=10m/s2, the total time of flight of the bullet is?

Answer 1

Vo = 200m/s[60o],
Yo = 200*sin60 = 173.2 m/s. = Ver. component of initial velocity.

Y = Yo + g*Tr = 0,
173.2 + (-10)Tr = 0,
Tr = 17.3 s. = Rise time.

Tf = Tr = 17.3 s. = Fall time.

Tr + Tf = Time in flight.