## krystal38garcia87 3 years ago 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

1. krystal38garcia87

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.

2. mth3v4

isnt this a math question?

3. jrodriguez2315

nope... university physics

4. mth3v4

ok did not see word gravity

5. krystal38garcia87

no help ? :(

6. mth3v4

looking im trying to figure out

7. krystal38garcia87

ok. thankyou

8. mth3v4

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

9. krystal38garcia87

huh?

10. mth3v4

it got too crowded up

11. Jemurray3

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.

12. krystal38garcia87

hey eashmore

13. eashmore

14. krystal38garcia87

it is nt right :(

15. krystal38garcia87

uhm

16. Jemurray3

17. krystal38garcia87

im pulling it u one second.

18. krystal38garcia87

up*

19. Jemurray3

Well in your question it says minutes...

20. krystal38garcia87

are you good with vectors?

21. krystal38garcia87

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

22. eashmore

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

23. krystal38garcia87

can you help me with a math problem?