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If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?

  • math -

    That would be when the velocity is zero. That happens when the derivative
    80 - 32t, is zero. Therefore t = 2.5 seconds at that time, and the maximum height is
    80*2.5 - 16(2.5)^2 = 100 ft

  • math -

    what ever

  • math -

    Alternatively:

    If a ball is thrown directly upward with a velocity of 80ft/s, its height (in feet) after t seconds is given by y=80t-16t^2. What is the maximum height attained by the ball?

    The time to reach the maximum height and zero velocity derives from Vf = Vo - gt wherer Vf = the final velocity = 0, Vo = the initial velocity = 80fps, t = the time to zero velocity and g = the deleration due to gravity.

    Therefore, 0 = 80 - 32t making t = 2.5 sec.

    Then, from h = 80t-16t^2, h = 80(2.5) - 16(2.5)^2 = 200 - 100 = 100 ft.

  • math -

    thanks!

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