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March 30, 2017

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if a ball is thrown vertically upward with an initial velocity of 48 ft/s, then its height, s, after t seconds is given by s(t)= 48t-16t^2.
a) what is the velocity of the ball after 1 second?
b) when does the ball hit the ground?
c) with what velocity does the ball hit the ground?

  • calculus - ,

    a) evaluate using t=1
    b) solve for t when s=0
    c) using t from (b) evaluate s'(t)

  • calculus - ,

    since t = 2,
    i) at 0.5 seconds,
    y= (52(2.5)-16(2)^2)/0.5=-13.6
    ii) (52(2.1)-16(2)^2)/0.1=-12.8
    and just add the time to 2 and divide by time

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