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Math Help - Projectile motion (1)

  1. #1
    Senior Member
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    Projectile motion (1)

    An aeroplane flies horizontally at height , h in a straight line at a constant speed , u . When it is directly above a cannon on the ground , the cannon fires a shell which is supposed to hit the airplane . Ignore the air resistance , determine in terms of u,g and h the minimum muzzle speed of the shell , v.

    Firstly , i resolve the velocity into 2 parts (ie v_x and  v_y)

     <br />
v_x=u\cos \theta\Rightarrow v^2_x=u^2\cos^2 \theta<br />

     <br />
v^2_y=u^2+2as<br />

    =u^2\sin^2 \theta-2gh

    v=\sqrt{v^2_x+v^2_y}

    -2gh}
    " alt="
    =\sqrt{u^2-2gh}
    " />

    The error here is in that sign , why should it be positive .
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  2. #2
    Senior Member nikhil's Avatar
    Joined
    Jun 2008
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    Lightbulb

    plane is travelling in a straight horizontal line with velocity u
    let v be the initial velocity of cannon mounted at angle x
    resolving velocity we get its
    horizontal component=vcos(x)
    vertical component=vsin(x)
    since cannon ball hit the plane we have
    vcos(x)=u (this will remain constant) equation 1
    since we have to calculate the minimum cannon ball velocity so the final vertical component of velocity of cannon ball will be 0
    therefor
    (vsin(x))^2=2gh equation 2
    squaring and adding equation 1 and 2 we have
    (vcos(x))^2+(vsin(x))^2=u^2+2gh
    therefor
    v^2=u^2+2gh
    and finally
    v=(u^2+2gh)^(1/2)
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