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1D particle motion help watch

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    A straight road on horizontal ground passes through a point ) in the direction e1 (due north). Let e2 point west and e3 point vertically upwards.
    An armed robber in a getaway car travels with velocity Ue1 along the road. At t=0, when the car passes the origin, the robber shoots a bullet at a distant police helicopter that is hovering above the ground stationary at Xe1 + Ye2 + Ze3, where X, Y and Z>0.
    The bullet trabels at velocity V cos theta e2 + (V sin theta - gt)e3 relative to the car, g=acceleration due to gravity and V>0, 0<theta<pie/2.

    a. Find x(t), the path of the bullet.

    b. Assuming the bullet hits the helicopter at time T, show that
    T= X/U and tan theta = Z/Y + gXsquared/2UsquaredY

    Any help would be appreciated. Thanks
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    I fail to see how there is enough information there for part (a). There are no vector quantities or angles or anything with actual values. The only thing you could do is give an answer in terms of V, X, Y, Z and θ probably. Maybe I'm missing something, maybe you're missing something, you sure you haven't missed anything out?

    And how is this 1D? There seem to be three dimensions.

    All I've been able to do is draw a pretty 3D picture with a car and helicopter.
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    (Original post by Joe_87)
    I fail to see how there is enough information there for part (a). There are no vector quantities or angles or anything with actual values.
    There are vector quantities.
    (Original post by Joe_87)
    or anything with actual values. The only thing you could do is give an answer in terms of V, X, Y, Z and θ probably.
    Why would you want to put in numbers? Isn't it much more useful to work out it for a general collection of parameters, then put in the parameter values at the end? Saves you working out the equations every time you have a new set of values for your parameters.

    In degree maths that is all you ever do. It is always general points (x,y,z) with general velocities (\dot{x},\dot{y},\dot{z}) and you have to develop the system of equations from that. At no point are actual numbers for the parameters involved.

    Initially the bullet starts at the origin (0,0,0) and its initial velocity is the velocity of the car plus the velocity given to it by the gun itself.

    Therefore the bullet's velocity is (U,V\cos \theta,V\sin \theta -gt)

    a) can be worked out simply by integrating the velocity with respect to time and using the initial condiction x(0)=(0,0,0).

    b) You have the x(t) equation, and you are told x(T)=(X,Y,Z). Equate terms in your x(T) expression to get the answers.
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    Well yeah clearly I don't know what I'm on about, but seeing as I'm only doing A level maths at the moment that would explain it.

    At least I tried to help the guy, nobody had replied to this after two hours.
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    sorry i was away but yeh this is degree level the problem was that i went to uni without having done mechanics i did statistics and i was told this was a level standard. Sorry for not having responded for such a long time.

    Thanks for all your help though im gna try that now
 
 
 
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Updated: November 16, 2005

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