Asked by Omen Stella
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?
Answers
Answered by
henry2,
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.
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.
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