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Minimum initial velocity

  1. Aug 28, 2015 #1
    1. The problem statement, all variables and given/known data
    At which minimum velocity should you throw the ball horizontally if you are standing on a hemispherical rock of radius R so that it at no point touches the rock and lands at the minimum distance from the rock horizontally. Find the expression that solves for initial velocity and woth that velocity calculate the distance traveled from the rock by the ball.

    2. Relevant equations
    X=Xi + Vcosu(t)
    Y=Yi + Vsinu(t) - g/2 (t^2)
    Vx=Vcosu
    Vy=Vsinu - gt

    3. The attempt at a solution
    It is definatelly clear that yf= 0 and yi=R
    From there t=x/V and 0=R-(1/2)gt^2 i gey R=(1/2)g(x^2/V^2), where to fo from here? x cant be R its path is oarabolic and would hit the rock if set on R and v is also unknown. It seems that there must be another correlation between y, x and R that i could use, but i dont see it.
     
  2. jcsd
  3. Aug 28, 2015 #2

    RUber

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    Can you define a path tracing the rock, and then write y_{ball} - y_{rock} > 0 for t< t_final?
     
  4. Aug 28, 2015 #3

    RUber

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    I get something that looks like this:
    Rock is defined by ##x^2 +y^2 =R^2## which gives ##y_{rock} = \sqrt{R^2 - x^2 } ##
    The path in x of the ball is defined by velocity and time, as in ##x = vt## initial trajectory is horizontal.
    The path in y of the ball is defined by R and time. ## y = R - 4.9t^2 ##.
    You want to find a velocity such that the only solution to y - y_{rock} = 0 is t = 0 (or possibly the t_{final}).
     
  5. Aug 28, 2015 #4
    I used the y(ball)^2 + x^2 > R^2 and got to the equation i posted below. Should i now solve for v?
     
  6. Aug 28, 2015 #5

    RUber

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    That should work. I don't see your equation, but I imagine it looks like:
    ## (R-4.9t^2)^2 + (vt)^2 > R^2 ##
    Which should simplify down to something like
    ## t^2 ( t^2 + f(v)) > 0 ## for all t, which means ##f(v)##>0.
    So, solve f(v) = 0 for the minimizer.
     
  7. Aug 28, 2015 #6
    Yes that is the equation but i dont know how you got f(v) there and what minimizer is.
     
  8. Aug 28, 2015 #7

    RUber

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    f(v) is just a placeholder for some function that depends on v, in this case is should be f(v) = v^2 - 9.8R.
    You are looking for the smallest velocity such that the inequality is true, so you are looking for the minimizer of the inequality.
     
  9. Aug 28, 2015 #8
    So its (gR)^(1/2) and the distance from the rock at the point of fall is 0 = R - (1/2)g(x^2/gR) so Δx= R( √2 + 1)
     
  10. Aug 28, 2015 #9

    RUber

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    Where did the +1 in your distance for x come from?
     
  11. Aug 29, 2015 #10
    It should be -1
     
  12. Aug 29, 2015 #11

    RUber

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    I'm still not seeing how you are coming up with that. What are you solving and how are you solving it?

     
  13. Aug 29, 2015 #12
    I want to know how far off the rock the ball is horizontally. I replace V^2 by gR in equation 0=R - 1/2g(x^2/gR) solve for x and calculate the Δx quantity to be Δx=x-R
    From above x=√2R so √2R - R = R(√2 - 1)
     
  14. Aug 29, 2015 #13
    Are we not looking for a parabola whose second derivative is the same at (0,R) as a circle centered at (0,0) with radius R?

    This also gives the distance the ball hits from the rock as R(sqrt(2)-1).
     
    Last edited: Aug 29, 2015
  15. Aug 29, 2015 #14
    Could you explain a little further how exactly does second derivative of a parabola being the same as that of circle at the point (0,R) correlates to the problem. How does one find the function if he knows f''parabola = f''circle located at point (0,0) with radius R at point (0,R). I have never met anything similar so im unfamiliar.
     
  16. Aug 29, 2015 #15

    RUber

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    Oh, I see. The question was asking for the horizontal distance from the rock. Sorry for not understanding before.
    I think you have this problem fully solved.
    There are many other ways to attack this problem, one of which is what Insightful mentioned. However, you have been able to solve it with the tools you have.
    Play around with the problem is a few ways, and you might find more relationships that will have to be true.
     
  17. Aug 29, 2015 #16
    It's simply that the curvature of the parabola at the top of the dome must equal the curvature of the dome. The slope of each (f') is obviously zero at that point, so the rates at which the slopes are changing (f") being equal seemed to me a reasonable approach. I can't be more specific because it just came to me while sketching circles under parabolas.

    Y1=sqrt(R2-x2)
    Y2=-Ax2+R
    Y2'=-2Ax
    Y2"=-2A

    so then get Y1" (a little messy, so not shown here) and set equal to -2A at x=0 to get A=1/(2R) and
    Y2=-x2/(2R)+R as the equation of the parabola.
     
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