## amrit_srovar 3 years ago A plane flies with a speed of 278 mps along a course which passes over an anti-aircraft gun on the ground. The airplane is at an altitude of 5390 m. If the muzzle velocity of the projectile is 735 mps with a slope of 4 vertical to 3 horizontal, determine the angle between the horizontal and the line of sight at which the projectile must be fired in order to hit the airplane during the upward motion of the projectile.

1. Fellowroot

Is it meters per second or miles per second?

2. amrit_srovar

meters per second

3. Fellowroot

I know you may not like this, but I'm going to say that the bullet never touches the aircraft. I may off on something, but I think I'm right.

4. amrit_srovar

is ur conclusion based on any solution or....?

5. Fellowroot

6. Fellowroot

I think I see where I'm wrong. It has to do with the bullet velocity. I'll redo.

7. amrit_srovar

ok

8. Fellowroot

I need to know what this muzzle velocity of the projectile is 735 mps is?

9. amrit_srovar

the muzzle velocity is the velocity of the projectile of the bullet as it leaves the muzzle of the gun. it is 735 mps

10. Fellowroot

Is it a horizontal component or vertical component or both or what. It tells me the slope but you can't use it directly because its probably a reduced form the slope. Slope is velocity and they say 4 up and 3 right, but you can't take those values literally. 8/6 is same as 4/3 so we don't know.

11. Fellowroot

Is it 735 in the vertical or horizontal? That's what I need to know.

12. amrit_srovar

well, i think it is both the horizontal and vertical v.

13. Fellowroot

and thats the problem.

14. amrit_srovar

but i think that we can get the horizontal and vertical v though the slope. use the slope to get the angle, then after u get the angle, use the 735 mps and the angle to get the vx and vy...im not sure with that

15. amrit_srovar

through*

16. Fellowroot

i think i got it hang on

17. Algebraic!

I looked at this question earlier. Honestly, it doesn't really make sense. They stipulate the angle and then they ask for the angle. Make sure you got the wording right, or, if there's a diagram that somehow explains this, post it.

18. amrit_srovar

well, i agree with u @Algebraic! because u can simply get the angle with the given slope. im also confused with that, but maybe the point is that we will use the angle to get the vx and vy, and then use the equations to get the theta

19. Algebraic!

impossible.

20. Algebraic!

changing the angle changes Vx and Vy.

21. Algebraic!

yet they stipulate Vy/Vx ... doesn't make sense.

22. Algebraic!

is this from an online problem set? or if it's from a text book, can you scan and post the pic. or take an ss?

23. amrit_srovar

it is from a book, but there's no picture. i posted exactly the same problem as it is in the book

24. Algebraic!

Well, I've tried interpreting this problem statement in a way that somehow makes sense.. but nothing has come to me.

25. amrit_srovar

or maybe we won't be using the slope....haha, i just don't know

26. amrit_srovar

27. Algebraic!

how?

28. amrit_srovar

this is our homework to be submitted tomorrow....=(

29. Fellowroot

|dw:1349577709975:dw|

30. amrit_srovar

=( i just don't know what to do

31. Fellowroot

|dw:1349578270124:dw|

32. Fellowroot

and i dont think the speed of the plane matters at all

33. Arindam

the angle between the muzzle and horizontal plane is tan-1(4/3)or 53.13 roughly. thus the initial vertical velocity of the missile is 735*Sin(53.13) or 588 m and accn is 9.8 mps^2 towards the ground. thus using the formula s = ut-.5gt^2 the time taken to reach 5390 m comes to be 10 secs. In this 10 secs the projectile moved horizontally 10*735*cos(53.13) metres or 4410 metres. and the plane has moved 2780 metres. Thus the plane was initially at (4410-2780) or 1630 metres horizontal distance from muzzle. thus the initial angle becomes tan-1(5390/1630) m or 73.174 m.

34. amrit_srovar

thnx a lot =)))))))))))

35. Fellowroot

How can the angle be arctan(4/3) =53 and then at the end its 73 ???