which one is angle will give the longest range ? 30 degrees ? 45 degrees ? 60 degrees ?

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asked by cj
  1. It depends on the initial height. However, if initial height = 0, then 45 degrees will give you the longest distance.

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    posted by Max
  2. range= VcosTheta*time
    hf=ho+VisinTheta*t-1/2 g t^2

    or 1/2 gt^2-ViSinTheta*t+(hf-ho)=0
    and you solve for t. Clearly, the hf-ho term at first glance seems to matter. Lets check it.

    using the quadratic equation.

    t= (ViSinTheta+-sqrt(Vi^2Sin^2Theta-2g(hf-ho))/g

    taking the + sqrt solution, put that into the horizontal equation..

    range=VicosTheta(ViSinTheta/g +sqrt(Vi^2sin^2theta-2(hf-ho)/g)

    now,with the assistance of calculus, maximizing range (drange/dtheta =0

    0=Vi^2 [sin^2theta/g - cos^2theta/g)+1/2 1/sqrt( ) *2sinthetacostheta)

    and you solve for theta

    and the solution is a lot of algebra, but what I want to point out, the factor (Hf-ho) is in that squareroot function in the denominator,so as Max points out, it matters the difference in height.

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