Vo = 1860m/s[67.4o]
Xo = 1860*Cos67.4 = 714.8 m/s.
Yo = 1860*sin67.4 = 1717.2 m/s.
Y = Yo-g*Tr = 0
1717.2 - 9.8Tr = 0
9.8Tr = 1717.2
Tr = 175 s. = Rise time.
Tf = Tr = 175 s. = Fall time
Tr+Tf = 175 + 175 = 350 s. = 5.83 Min. in flight.
An artillery shell is fired at an angle of 67.4◦ above the horizontal ground with an initial speed of 1860 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell, neglecting air resistance.
Answer in units of min
1 answer
1 answer