physics
posted by alexa .
An artillery shell is fired at an angle of 48.2 degrees above the horizontal ground with an initial speed of 1530 m/s. THe acceleration of gravity is 9.8 m/s^2. Find the total time and flight of the shell,neglecting air resistance.Answer in units of minutes.
Find its horizontal range,neglecting air resistance. Answer in units of km

Vo = 1530m/s @ 48.2 deg.
Vo(h) = 1530cos48.2 = 1020m/s.
Vo(v) = 1530sin48.2 = 1141m/s.
Vf = Vo(v) + gt,
Solve for t and get:
t = (Vf  Vo(v)) / g,
t = (0  = 1141) / 9.8 = 116.4s = t(up) = t(down).
T = 2 * 116.4 = 232.8s = 3.9 min. =
total flight time.
Hor Range = Vo(h) * T,
Hor Range = 1020m/s * 232.8s = 237,456m
= 237.5km.