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Physics

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The problem contains graphics that I don't know how to show other than in an image format:

h t t p : / / b i t . l y / c D H P V M
h t t p : / / b i t . l y / 9 m b y C R

  • Physics - ,

    And I still haven't gotten an answer for my previous question here:

    h t t p : / / b i t . l y / d g 6 o O F

  • Physics - ,

    I can't make the "previous" graphic open.

    On this problem, label the horizontal distance H (dcosPHI) and the final height yf(dsinPHI)

    horizontal:
    VcosTheta*time=x=dcosPHI
    vertical
    yf=VsinTheta*time-1/2 g t^2

    from the first equation
    time=dcosPHI/VcosTheta
    and putting this in the second equation
    dsinPHI=VsinTheta*dcosPHI/VcosTheta-1/2 g (dcosPHI/VcosTheta)^2

    multiplying both sides by (VcosTheta)^2
    dsinPHI(VcosTheta)^2=V^2 sinThetacostheta d cosPHI=1/2 g dcosPHI

    so take the derivative of d with respect to theta, set to zero, solve for theta in terms of V and PHI.

    A bit of algebra is required, so get a large pad of paper.

  • Physics - ,

    What does d= ?

    (Not sure what to take the derivative of)

  • Physics - ,

    d is the height up the hill, as drawn on your figure. As I recall, you are trying to maximize d, so d d/dTheta=0=..

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