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Homework Help: Minimum Velocity needed to clear a hemisphere

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a hemisphere of radius R metres on the ground. You are standing at the very top of the hemisphere and you kick a ball out such that it travels outwards horizontally with speed v metres per second. During the trajectory of the ball, it does not come into contact with the hemisphere at all.
    a) What is the minimum v that can fulfil the above conditions?
    b) How far away from the centre of the hemisphere will the ball land?

    2. Relevant equations
    Projectile Motion Equations


    3. The attempt at a solution

    Looking at the y-axis : vt = L, where L is the horizontal distance travelled after time t
    Looking at the y-axis : H = 0.5*g*t^2, where H is the vertical distance from the top of the hemisphere after time t

    Therefore, at time t, the distance of the ball above the ground is R - 0.5*g*t^2.

    At the point that is x metres away from the centre, the height of the slice of hemisphere at that point is sqrt(R^2 - x^2).

    I can substitute x with vt and get sqrt(R^2 - v^2*t^2)

    So now, in order to fulfil the conditions:

    R - 0.5*g*t^2 > sqrt(R^2 - v^2*t^2) for t = 0 to the time the ball reaches the ground

    Squaring both sides, I get:

    R^2 - R*g^2*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2 (I think I dont have to change the bigger-than sign to smaller-than because I know that for this range of values, both sides>0)

    So if I make v the subject:

    v^2*t^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4
    v^2 > R^2 + R*g^2*t^2 - 0.25*g^2*t^4/t^2
    v > sqrt(R^2 + R*g^2*t^2 - 0.25*g^2*t^4)/t

    But I still have the t term on the Right Hand Side. How do I get rid of it, or is there another way to do this question?

    Thanks all in advance.
     
  2. jcsd
  3. Feb 11, 2010 #2
    Can someone please help me? Thanks
     
  4. Feb 11, 2010 #3

    ehild

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    Homework Helper



    there is a mistake: g is not squared:

    R^2 - R*g*t^2 + 0.25*g^2*t^4 > R^2 - v^2*t^2

    You can subtract R^2 from both sides:

    - R*g*t^2 + 0.25*g^2*t^4 > - v^2*t^2


    The inequality has to hold for all t, even at t=0. So v^2>R*g. If v is less, the ball rolls downhill.

    ehild
     
  5. Feb 11, 2010 #4
    Thank you very much! :)
     
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