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What is the speed of bomber when it releases its cargo

  1. Feb 26, 2017 #1
    1. The problem statement, all variables and given/known data
    During a drop, a bomber decends with velocity v, directed at angle 37 degrees below horizontal. The cargo is released at an altitude of 2450 m and reaches intended dropzone with displacement of 3850 m

    What is the speed of bomber when it releases its cargo
    2. Relevant equations
    [itex]
    V_{0x} + a_x t = V_x ----


    \Delta x = (1/2)(V_{0x} + V_x) t ----


    \Delta x = V_{0x}t + (1/2)a_x t^2 -----


    \Delta x = V_x t - (1/2) a_x t^2
    [/itex]

    3. The attempt at a solution
    So I begin with a triangle.

    Hypotenus = 3850 m. angle opposite hypotenus is 90 degrees. I find the top angle to be 53 degrees, because the plane is coming down at angle 37 degrees. That leaves angle at the drop to be 37 degrees.

    Does everything seem right so far?

    The length of the 37 degrees angle is 2450 m - x, but I find it using law of sines.

    3850 m / sin(90) = x / sin(37)

    I find it to be 2316 m, and using pythagoras I get 3075 m to be the length for angle 53 degrees.

    Now, I'm being asked to find v, speed at which plane was going as cargo is dropped.

    [itex] V = \sqrt {V_{0x}^2 + V_{0y}^2 } [/itex]

    Now this is my second attempt at this problem and I'm wondering why this method doesnt make sense. I will also post my first method and see if its okay with y'all, I think it is but it just seems like sooo much work.

    Since [itex] V_{0x} = V_x [/itex], I use [itex] \Delta x = (1/2) (V_{0x} + V_x) t [/itex] and get

    [itex] V_{0x} = \Delta x /t [/itex]

    For [itex] V_{0y} = xsin(37) [/itex]

    I don't see anything wrong with the statements I made so far. If anyone does please let me know here.

    Anyways, [itex] V_{0x} = xcos(37) = \Delta x /t [/itex]
    so [itex] x = \Delta x / t cos(37) [/itex]

    Now plugging x into [itex] V_{0y} [/itex]

    [itex] \Delta x tan(37) / t [/itex] this = [itex] V_{0y} [/itex]

    Now I want to use equation [itex] \Delta y = V_{0y} t + (1/2) a_y t^2 [/itex]

    [itex] \Delta y = \Delta x tan(37) + (1/2) a_y t^2 [/itex]

    Notice how I got rid of the t in the original motion equation. Then I solved:

    [itex] \sqrt {2(\Delta y - \Delta x tan(37))/a_y} = t [/itex]

    But the value that I get for t using this method just doesn't make sense. It's under 1, and no way can the cargo be dropped off in under 1 second.

    I'm about to make a post in the comments with my other method which was much uglier, but I think would provide a correct answer... , but can anyone explain to me why what I did here doesn't make sense?
     
  2. jcsd
  3. Feb 26, 2017 #2
    I had a very difficult time understanding your work. Can you tell me what you calculated for the horizontal distance that the bomb traveled?
     
  4. Feb 27, 2017 #3
    3075 m.

    I find the top angle to be 53 degrees, because the plane is coming down at angle 37 degrees.

    Using law of sines with 3850m/sin(90) = x/sin(53) gave me x = 3075 m
     
  5. Feb 27, 2017 #4
    You know that the displacement is 3850. And you know that the height is 2450 m. So from that alone you can calculate the horizontal distance that it traveled. The angle of that triangle is not 53°. The initial angle of the bomb starts out as 53°, but it changes continually as the bomb falls.

    Edit: I should have said that the angle of displacement is not 53°. It starts out as 53° but changes continually - getting smaller and smaller - as the bomb continues on its path.

    Edit 2: One other comment: It looks like you are using x as the variable for initial velocity. If that is true, x is a very poor choice since you are using Δx for distance. Why not use Vo - especially since you are already using Vox and Voy.
     
  6. Feb 27, 2017 #5

    ehild

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    Make a drawing and show that triangle. I think you want to find the maximum displacement in the horizontal direction, but you use the wrong angles. It is the initial velocity that makes the angle 37° below the horizontal.

    No. See the picture. The red arrow represents the initial velocity of the bomb, the same as the velocity of the plane. The blue line is the trajectory of the bomb. Δx can be found from the light blue triangle using Pythagoras.
    upload_2017-2-27_7-39-42.png
     
  7. Feb 27, 2017 #6
    That picture is worth a thousand words.
     
  8. Feb 27, 2017 #7

    ehild

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    Thanks :oldsmile:
     
  9. Feb 27, 2017 #8
    Sorry everyone I forgot to note important information in the OP.

    The question is:
    A ww2 era dive bomber is being used to drop food and supplies to outposts on remote mountaintops. During one such drop, the bomber descends with velocity v directed at an angle of 37 degrees below the horizontal. The cargo is released at an altitude of 2450 m and reaches intended dropzone with a displacement of 3850 m. What is the speed of the bomber when it releases its cargo?

    So since the bomber is on a mountain top, but altitude means from sea level right, so the [itex] \Delta y [/itex] must be 2450 m - x, correct?

    http://i.imgur.com/otPRfTm.png
    What can I do with the given information to complete my triangle? So far I have one angle, 90 degrees, and its side is 3850 m long, then I have two more sides, [itex] \Delta y [/itex] and [itex] \Delta x [/itex] and their angles are uknown. How do I proceed from here?
     
    Last edited: Feb 27, 2017
  10. Feb 27, 2017 #9

    haruspex

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    Since that would leave you with inadequate information, you should take it as meaning relative to the mountaintop.
     
  11. Feb 27, 2017 #10
    Thanks for the info. It led to me obtaining the answer. Wow this question was not half as complicated as I thought it would be.

    Posting my solution for anyone else following this problem:

    So you get a triangle with hypotenus = 3850 m, [itex] \Delta y = 2450 m [/itex], [itex] \Delta x = 2969.848481 [/itex]

    [itex] V_x = \Delta x/t [/itex]

    [itex] (\Delta y-(1/2)a_yt^2)/t = V_{0y} [/itex]

    [itex] x = V_x/cos(37), x = V_{0y}/sin(37) [/itex]

    set x = to each other

    [itex] \Delta x/tcos(37) = (2\Delta y - a_yt^2)/2tsin(37) [/itex]

    solving for t

    [itex] 2\Delta x tan(37) = 2 \Delta y - a_y t^2 [/itex]

    solve for t and you get 6.578546595s, plug back in for

    [itex] V_x = \Delta x/t [/itex]

    [itex] (\Delta y-(1/2)a_yt^2)/t = V_{0y} [/itex]

    and then find [itex] \sqrt {v_x^2 + v_{0y}^2 } [/itex] and you get answer 566 m/s

    thanks Tomhart, ehild, and gneill for the help.. If I could ask one last thing is for you guys to tell me if my method is sound?
     
  12. Feb 27, 2017 #11
    I think your method looks right. In the future, please try to make the following substitutions: V for velocity instead of x, and haruspex in place of gneill. :)
     
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