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

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- anonymous

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.

- anonymous

isnt this a math question?

- anonymous

nope... university physics

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## More answers

- anonymous

ok did not see word gravity

- anonymous

no help ? :(

- anonymous

looking
im trying to figure out

- anonymous

ok. thankyou

- anonymous

lol
i can only see the first part of what you typed
the rest i got lost

- anonymous

huh?

- anonymous

it got too crowded up

- anonymous

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.

- anonymous

hey eashmore

- anonymous

kyrstal. Didn't I answer this question for you already?

- anonymous

it is nt right :(

- anonymous

uhm

- anonymous

You need to watch your units. Your answer should be in minutes, not seconds, right?

- anonymous

im pulling it u one second.

- anonymous

up*

- anonymous

Well in your question it says minutes...

- anonymous

are you good with vectors?

- anonymous

it was right :/ just didnt read the whole problem.

- anonymous

The work you did was correct. As Jemurray3 correctly pointed out, you need to convert seconds to minutes.

- anonymous

can you help me with a math problem?

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