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 75.7 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 32.2◦ above the...
- physics - A shell fired from the ground with an initial speed of 1.70 x 10^3 m/...
- physics - A shell is fired from the ground with an initial speed of 1.54 103 m/s...

More Related Questions