Finding the angle of release from a moving plane that drops an object X distance away

  1. 1. The problem statement, all variables and given/known data

    A plane flies horizontally @ 299 m/s relative to ground.
    Altitude: 3450m, level terrain; neglect air resistance.

    1: How far from the point vertically under the point of release does the object land (in meters)?

    2: At what angle must the bomb be released so that it hits the target at time of release (in °)?

    2. Relevant equations

    t = (sqrt(Voy^2 + 2gh) - Voy) / -g

    Range = Vox(t)

    3. The attempt at a solution

    Using the above, I correctly figured 7933.83 m for the range. It's the angle part that's screwing me.

    Now, because the y-component of that object's velocity is zero at release, I was able to answer #1 without breaking anything down. But I have not a clue as to how to find that damn angle.
  2. jcsd
  3. Doc Al

    Staff: Mentor

    I assume they want the bomb to hit the target nose first. What angle does the velocity make when it hits the target? (Find the velocity components.)
  4. I'm really not sure what I'm doing on this one. Would you mind starting me off a bit?
  5. nrqed

    nrqed 3,101
    Science Advisor
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    Gold Member

    To me the second question does not make sense. First, the bomb cannot hit the target at the time of release! Second, I have the *feeling* that they mean to ask the angle at which it hits the target, not at which it is released, but I can't tell for sure. Third, if they want an angle of release, they have to give more information (how far the target is when th ebomb is released and more stuff as well). is this the way is exactly phrased??
  6. It really is worded like that. "At what angle from the vertical at the point of release must the telescopic bomb sight be set so that the bomb hits the target ... ?" The distance to target is 7933.83 m as figured from the first half of the problem, and the rest of the info is in my post. I'm just having a hard time figuring out what to use and where.
  7. Dick

    Dick 25,913
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    So now you have right triangle composed of a vertical line from the plane to a point directly below on the ground and a horizontal line from that point to the target. The bomb sight points from the plane to the target along the hypotenuse. You know the length of both of these sides and want to find the angle at the corner where the plane is. Sounds like trig to me.
    Last edited: Jun 14, 2007
  8. nrqed

    nrqed 3,101
    Science Advisor
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    Gold Member

    AHHH!!!!!! It makes a HUGE difefrence to mention the telescopic bomb sight!!!:rolleyes:
    You just have to find the angle at which the target appears from the bomb's point of view when the bomb is released. Dick explained it. It has nothing to do with physics at this point, it's just trigonometry.
  9. Yeah, sorry for screwing up the original question by trying to over-Cliffs Notes it.

    tan^-1 (7933.8264 / 3450) = 66.498° it is.

    Thanks. o:)
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