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    • Thread Starter

    Hi i am on the last question of my hw and i am stuck and i was wondering if anyone could help me on it

    Q8) A ball is thrown vertically upwards with a speed of 15 ms-1 from a point 1m above the ground. Find the speed with which it hits the floor .If it rebounds with a speed which is half the speed with which it hits the floor, find its greatest height after the first bounce.

    I am completely stumped here could ANYONE help me on this question:confused:

    For part a)
    s=-1\\u=0\\v=?\\a=-g \\t=?

    Use: v^2=u^2+2as

    And then for part b)

    Your new u is half your answer for a).
    v=0 \\ a=-g

    Use:  v^2=u^2+2as to find s, the highest height (eg. the height when velocity = 0)

    We start at s=0, so the height of the ground is -1.
    The initial speed, u, is 15, and the acceleration is -g (take upwards as positive).
    So v=sqrt(15^2 - 2g(-1)).
    Your answer will depend on g, usually we take g=10, 9.8 or 9.81.
    Let the rebound speed be w, so w=0.5v.
    At the greatest height, the velocity will be momentarily zero (think about it).
    So using the same formula:
    0^2 = w^2 -2gh
    So h=(w^2)/2g
    • Thread Starter

    cheers guys u really helped me out thanks for the help its gr8ly appreciated
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Updated: October 16, 2006

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