Posted by **alexa** on Thursday, June 23, 2011 at 12:18am.

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

- physics -
**Henry**, Thursday, June 23, 2011 at 4:45pm
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.

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