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.
Answer This Question
Related Questions
 physics  An artillery shell is fired at an angle of 66.1 degrees above the ...
 Physics  An artillery shell is fired at an angle of 67.4◦ above the ...
 physics  An artillery shell is fired at an angle of 86◦ above the ...
 physics  An artillery shell is fired at an angle of 45.3◦ above the ...
 Physics  An artillery shell is fired at an angle of 67.4◦ above the ...
 Physics  An artillery shell is fired at an angle of 26.8 degree above the ...
 Physics  An artillery shell is fired at an angle of 55.7◦ above the ...
 Physics  An artillery shell is ﬁred at an angle of 75.7 above the ...
 Physics  An artillery shell is ﬁred at an angle of 32.2◦ above the...
 physics  A shell fired from the ground with an initial speed of 1.70 x 10^3 m/s...
More Related Questions