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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

Physics
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vertical is 1520sin62.7 = 1350.69819 horizontal is 1520cos62.7 = 697.14732 s=ut +1/2 at^2 therefore 1350.698t +4.9t^2 t=277.77 seconds... 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.
isnt this a math question?
nope... university physics

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Other answers:

ok did not see word gravity
no help ? :(
looking im trying to figure out
ok. thankyou
lol i can only see the first part of what you typed the rest i got lost
huh?
it got too crowded up
Assuming you plugged the values correctly into your calculator, that's correct. The procedure you went through is right. You need to convert from seconds to minutes, though. For part b, I assume you're looking for the total range of the projectile? In which case you need multiply the time of flight (in seconds) by the velocity.
hey eashmore
kyrstal. Didn't I answer this question for you already?
it is nt right :(
uhm
You need to watch your units. Your answer should be in minutes, not seconds, right?
im pulling it u one second.
up*
Well in your question it says minutes...
are you good with vectors?
it was right :/ just didnt read the whole problem.
The work you did was correct. As Jemurray3 correctly pointed out, you need to convert seconds to minutes.
can you help me with a math problem?

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