Monday, November 30, 2009

Fox 197

Those apes keep juglin'


  1. found 4sqrt(1+a/b)
    maybe there is mistakes but the path is right i think

  2. Let the velocity at the top left corner of the wall be V and the angle made by V to the horizontal be θ. Horizontal range=b=V^2sin(2θ)/g. For smallest possible initial effort, 2θ=90 or θ=45 deg. which gives V^2=b*g and the horizontal component of V = Vx = V(bg/2). If the initial velocity of the throw is U then U^2=V^2+2ag=bg+2ag and Ux=[(bg+2ag)^(1/2)]*cos(λ). But Ux=Vx. Hence,(bg+2ag)^(1/2)*cos(λ)=V(bg/2) or cos(λ) = V(b/(4a+2b))& sec(λ)= V(4a+2b)/b). We know that (sec(λ))^2=1+(tan(λ))^2 or (4a+2b)/b=1+(tan(λ))^2 or tan(λ)=V((4a+b)/b) =V(1+4a/b). QED.

  3. Nice work! Thank you.

    A few observations:
    a=0 => tan(λ)=1 => λ=45
    b/a=2 => tan(λ)=sqrt(3) => λ=60, which was the case in Fox 193.
    a>0 and b=0 => λ=90.

    All makes sense.