anonymous
  • anonymous
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.
Physics
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions.

anonymous
  • anonymous
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.
Physics
katieb
  • katieb
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

Fellowroot
  • Fellowroot
Is it meters per second or miles per second?
anonymous
  • anonymous
meters per second
Fellowroot
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
is ur conclusion based on any solution or....?
Fellowroot
  • Fellowroot
1 Attachment
Fellowroot
  • Fellowroot
I think I see where I'm wrong. It has to do with the bullet velocity. I'll redo.
anonymous
  • anonymous
ok
Fellowroot
  • Fellowroot
I need to know what this muzzle velocity of the projectile is 735 mps is?
anonymous
  • anonymous
the muzzle velocity is the velocity of the projectile of the bullet as it leaves the muzzle of the gun. it is 735 mps
Fellowroot
  • 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.
Fellowroot
  • Fellowroot
Is it 735 in the vertical or horizontal? That's what I need to know.
anonymous
  • anonymous
well, i think it is both the horizontal and vertical v.
Fellowroot
  • Fellowroot
and thats the problem.
anonymous
  • anonymous
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
anonymous
  • anonymous
through*
Fellowroot
  • Fellowroot
i think i got it hang on
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
impossible.
anonymous
  • anonymous
changing the angle changes Vx and Vy.
anonymous
  • anonymous
yet they stipulate Vy/Vx ... doesn't make sense.
anonymous
  • anonymous
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?
anonymous
  • anonymous
it is from a book, but there's no picture. i posted exactly the same problem as it is in the book
anonymous
  • anonymous
Well, I've tried interpreting this problem statement in a way that somehow makes sense.. but nothing has come to me.
anonymous
  • anonymous
or maybe we won't be using the slope....haha, i just don't know
anonymous
  • anonymous
please help =(
anonymous
  • anonymous
how?
anonymous
  • anonymous
this is our homework to be submitted tomorrow....=(
Fellowroot
  • Fellowroot
|dw:1349577709975:dw|
anonymous
  • anonymous
=( i just don't know what to do
Fellowroot
  • Fellowroot
|dw:1349578270124:dw|
Fellowroot
  • Fellowroot
and i dont think the speed of the plane matters at all
anonymous
  • anonymous
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.
anonymous
  • anonymous
thnx a lot =)))))))))))
Fellowroot
  • Fellowroot
How can the angle be arctan(4/3) =53 and then at the end its 73 ???

Looking for something else?

Not the answer you are looking for? Search for more explanations.