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    Hi, I think this is meant to be simple, but I have no clue how to do these questions. My teacher zoomed through this topic without explaining it properly.

    A cricketer throws a ball vertically upwards so that the ball leaves his hands at a speed of 25ms(-1). If Air resistence can be neglected, calculate:

    The maximum height reached by the ball

    The time taken to reach maximum height

    The speed of the ball when it is at 50% of the maximum height.

    For the first question, I have to work out S, and I only know V (and U?) and A (am I meant to use -9.81 here?).I 'm not sure wether I'm meant to use the constant accelaration equations, but I'm stumped.
    I would really appreciate a few hints

    The max height of a body is given by

    mgh = 0.5mv^2
    gh = 0.5v^2

    h = v^2/2g

    a=-9.8 m/s/s
    v=0 m/s
    u= 25 m/s

    v^2 = u^2 +2as
    0 = 625 -19.6S
    s= 31.89m

    For part two plug in the values above into s=ut + 1/2at^2

    For part 3 divide s (height) from part one by two, and then plug in that value as s, and all the other values apart from v into the original equation.
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Updated: October 14, 2009

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