Wednesday, February 24, 2010

Fox 253


The answer is now confirmed! Will post later...

5 comments:

  1. y=1/x & dy/dx=-1/x^2. So slope of the normal at the pt. of reboumd is = x^2 = 4 or tan(α)=4 or cos(α)=1/√17. Velocity along normal=(0.97014)√(2g*144.5). Eqn of parabolic path with the point of rebound as the origin: y = 4x - x^2/32 after substituting all relevant values. At x=64 the peak is reached with h = 128+0.5 = 128.5 above the original x-axis. Unfortunately, this option isn't there. So pl. tell me where I'm going wrong.
    Ajit

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  2. "So slope of the normal at the pt. of reboumd is = x^2 = 4"

    then the laungh angle should be (90 - 2*alpha). is that not the case?

    launch angle: angle btw the rebound and the horizontal.

    -binary

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  3. Launch angle is the angle between the initial velocity direction and the x-axis i.e. the direction of the normal & the x-axis in our case or tan(α)=4 so the launch angle is ~ 75.964 deg. to the horizontal and the general equation is:
    y = xtan(α)-(g/2)(x/ucos(α))^2 where the point of rebound is the origin.

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  4. i still think "tan(α)=4" is the slope of the normal => launch angle is 90 - 2α.
    -binary

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  5. sorry i think i made amistake. let me try again. tangent to 1/x at x=2 has a slope of m1=-1/4. slope of the normal=m2. m1*m2=-1 => m2=4. => slope of the normal is 4. let a be the launch angle. let b be the angle between normal and the y-axis. tan(b)=1/4. we also know:
    a+2b=90 => 2b=90-a
    => tan(2b)=tan(90-a)
    => 2tan(b)/(1-tan^2(b))=cot(a)
    => 2tan(b)/(1-tan^2(b))=cot(a)
    => 2(1/4)/(1-1/16)=cot(a)
    => 1/2 /(15/16)=cot(a)
    => 8/15 =cot(a) it is a 8-15-17 triangle!!!
    => tan(a)= 15/8
    => sin(a) = 15/17
    => cos(a) = 8/17
    so the launch angle has a cos of 8/17. rest should follow.
    -binary

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