Question 1190110: An artillery shell is fired with an initial velocity of 760 m/s at an angle of 25o elevation. Calculate the time of flight for the shell.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! An artillery shell is fired with an initial velocity of 760 m/s at an angle of 25o elevation. Calculate the time of flight for the shell.
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The vertical component is 760*sin(25) = ~ 321.2 m/sec
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Using 9.8 m/sec/sec for gravity, the time to apogee is 321.2/9.8 = 32.77 seconds
The time descending is the same.
Flight time = 2*32.77 = 65.54 seconds
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Seems like a long time but might be realistic for big guns, 155 mms, or 175 mms.
I had to deal with artillery while flying in Vietnam. It's good not to get hit by it. While flying or any time.
Once, on the way back from dropping a patient on a hospital ship about 15 miles offshore, I saw the USS New Jersey firing inland. 16 inch (diameter) guns, firing 2000 pound rounds with a 1 second rocket burn. Impacting 35 miles away inland. When 3 guns fired at the same time, a wake was generated on the port side of the ship. It backed up sideways. Ft = mv.
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