Question

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

Answer 1

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.

Answer 2

isnt this a math question?

Answer 3

nope... university physics

Answer 4

ok did not see word gravity

Answer 5

no help ? :(

Answer 6

looking
im trying to figure out

Answer 7

ok. thankyou

Answer 8

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

the rest i got lost

Answer 9

huh?

Answer 10

it got too crowded up

Answer 11

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.

Answer 12

hey eashmore

Answer 13

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

Answer 14

it is nt right :(

Answer 15

uhm

Answer 16

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

Answer 17

im pulling it u one second.

Answer 18

up*

Answer 19

Well in your question it says minutes...

Answer 20

are you good with vectors?

Answer 21

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

Answer 22

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

Answer 23

can you help me with a math problem?