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Minimum distance between bomb and shell

  1. Aug 2, 2016 #1
    1. Prolem Statement:-
    An aircraft is flying horizontally with constant velocity = ##200 m/s##, at a height = ##1 km## above the ground. At the moment shown, a bomb is released from aircraft and the canon- gun below fires a shell with initial speed = ##200 m/s##, at some angle #\theta#.
    1)For what value of ##'\theta'## will the projectile shell destroy the bomb in mid- air ? If the value of ##'\theta'## is ##53^{\circ}##.
    2)Find the minimum distance between the bomb and the shell as they fly past each other. Take ##\sin53^{\circ}=4/5##.

    The figure for the question is as shown below:
    2380 (2).png

    2. My solution:-

    I have solved the first part of the question the one that I am having trouble solving is the second part. Here is my work on the solution.
    For reference these are the variables that I have used in my solution:-
    • ##\vec{s}_{s/e}##-Displacement of shell w.r.t Earth
    • ##\vec{s}_{b/e}##-Displacement of bomb w.r.t Earth
    • ##\vec{v}_{0}##-Velocity of bomb/shell w.r.t Earth
    Lets not list all of them, I think you should have got the idea now. So, here goes my solution.

    ##\vec{s}_{s/e}=-(v_0\cos{\theta}t)\hat{i} + (v_0\sin{\theta}t-(1/2)gt^2)\hat{j}##
    ##\vec{s}_{b/e}=(v_0t)\hat{i}+(1-(1/2)gt^2)\hat{j}##

    Now, to find the minimum distance between the bomb and the shell what I did was first find the relative displacement and then differentiate its modulus and set it as ##0## which gives us the minimum possible distance between them.

    ##\begin{aligned}
    &\vec{s}_{s/b}=-(v_0\cos{\theta}+v_0)t\hat{i}+(v_0\sin\theta-1)\hat{j} \\
    \implies & |\vec{s}_{s/b}|^2=2v_0^2t^2(1+\cos\theta)-2v_0\sin\theta t \qquad\qquad\qquad\qquad\qquad \ldots(1)
    \end{aligned}##

    Now, differentiating equation ##(1)## w.r.t ##t## and putting ##\dfrac{d|\vec{s}_{s/b}|^2}{dt}=0##, we get

    ##t=\dfrac{\sin\theta}{2v_0(1+\cos\theta)}=\dfrac{1}{4v_0}##

    On substituting ##t=1/(4v_0)## in equation ##(1)##, we get

    ##{|\vec{s}_{s/b}|}^2=\dfrac{320}{16\times 25}=\dfrac{4}{5} \implies |\vec{s}_{s/b}|=\dfrac{2}{\sqrt5}##

    3.Book's Solution:-

    2380.png


    Why does my answer differ from the book, what did I miss.
     
    Last edited: Aug 2, 2016
  2. jcsd
  3. Aug 2, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    An image showing the initial relative positions of the aircraft and gun would be helpful. Your equation for the trajectory of the shell indicates that its x-component is in the direction of the negative x-axis, while that of the bomb is positive. If they both start at the same x-coordinate then surely they can't intersect as they would be moving in opposite directions away from the origin. So there must be some initial separation in the x-direction.
     
  4. Aug 2, 2016 #3
    I have been trying to upload for quite some time but don't know why it doesn't get uploaded.
     
  5. Aug 2, 2016 #4
    Finally, it got uploaded.
     
  6. Aug 2, 2016 #5

    gneill

    User Avatar

    Staff: Mentor

    So according to your diagram the initial x-separation is ##\sqrt{3}~km##. You'll have to work that into your position equations.
     
  7. Aug 2, 2016 #6
    Oh my god.....that was a blunder, thanks for all your efforts.
     
    Last edited: Aug 2, 2016
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