An artillery shell is fired at an angle of 62.7
above the horizontal ground with an initial
speed of 1520 m/s.
The acceleration of gravity is 9.8 m/s2 .
Find the total time of flight of the shell,
neglecting air resistance.
Answer in units of min
Stacey Warren - Expert brainly.com
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vertical is 1520sin62.7 = 1350.69819
horizontal is 1520cos62.7 = 697.14732
s=ut +1/2 at^2 therefore
and then divide 697.1432/277.77 = part b
Can someone tell me if this is the correct answer, or if not what I am doing wrong.
oh yeah this is part B lol... Find its horizontal range, neglecting air resis-
Answer in units of km
It appears you know what you're doing. Check this wikipedia article for an equation reference: http://en.wikipedia.org/wiki/Projectile_motion
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I always make small eroors in my calculations and I end up with the wrong answer. Did you by chance check the work to see if it is correct. I didnt want to enter it into my HW assignment until I know for sure it is correct.