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    mathematics mechanics help

    yutyy
    Last edited by longy; February 23rd 2008 at 01:11 PM.
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    Quote Originally Posted by longy View Post
    A ball is projected vertically upwards with an initial speed of 30ms^-1 the time taken for the ball to reach its maximum height is 3 s and the maximum height that the ball reaches is 30m.The remainder of the question we revise by taking air resistance in to account.Model the the ball as a shpere of radius D,mass m,and assuming that the quadratic model of air resistance applies.My question is i need to show that the acceleration at time t is a(t)i in which a(t)=-g/b^2(v(t)2+b^2) where v(t)i is the velocity of the ball at time t,and b^2=mg/0.3D^2 I f somebody could help me or give me a few tips thanks.
    The equation of motion will be given by the differential equation
    ma = mg - kv^2

    I would call the quadratic resistance coefficient b, but the units for it are all wrong. You'll have to figure out what to do with that yourself.

    So
    \frac{dv}{dt} = g - \left ( \frac{k}{m} \right )v^2

    Solve for v.

    -Dan
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