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Kinematics of a falling object to hit target

  1. Nov 17, 2015 #1
    1. The problem statement, all variables and given/known data
    An airplane flies horizontally with a speed of 300m/s at an altitude of 300m. Assume the ground is level. At what horizontal distance from a target must the pilot release a bomb as to hit the target?

    Answer: 2.7km

    2. Relevant equations
    a=v/t=d/t2
    r_f = r_i + v_i*t + 1/2at2

    3. The attempt at a solution
    I tried approaching this question using a kinematics equation and just the basic a=v/t approach...

    a=v/t=d/t2
    t = sqrt( 2 * d / g ) = 9s
    d(horizontal) = vt = (300)(9) = 2.7km

    r_f = r_i + v_i*t + 1/2at2
    Using the quadratic equation to find t = 62.5s
    d(horizontal) = (300)(62.5) = 18.7km

    Why can I not use the kinematics equation to approach this problem?
    In general, how do I know whether its appropriate to approach a problem using kinematics vs. the "basic" equations (a=v/t; v=d/t)?
     
  2. jcsd
  3. Nov 17, 2015 #2

    SteamKing

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    For one thing, the vertical velocity of the bomb is not constant, so how do you calculate a = v/t?

    And another thing. You haven't calculated the correct free-fall time of this bomb when it is dropped from an altitude of 300 m. Check your arithmetic again.
     
  4. Nov 17, 2015 #3
    A bit of a typo, but I meant that the altitude was 400m. Either way, you're right I wasn't calculating it correctly.
    First off, I realized that the initial velocity of the bomb is 0m/s. Just because its being dropped from a moving plane, free fall is independent of how it starts falling.

    r_f = r_i +v_i*t + 1/2at2
    0 = 400 - 4.9t2
    400=-4.9t2
    t = 9 seconds to for the bomb to reach the ground.

    300m/s * 9 seconds = 2.7km
    I multiplied these because the plane is going to move an "extra" 300m for every second.
     
  5. Nov 17, 2015 #4

    haruspex

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    More to the point, the bomb continues forwards at 300m each second.

    To answer your question about when you can use certain equations, the only general kinematics equations involve integrals and derivatives. The SUVAT equations can be deduced from those when acceleration is constant. That includes a=v/t (or more accurately, ##a=\Delta v/\Delta t##). But v=d/t is only valid when there is no acceleration.
     
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