calculus

If a ball is thrown vertically upward from the roof of 64 foot building with a velocity of 64 ft/sec, its height after t seconds is s(t)=64+64t–16t2 . What is the maximum height the ball reaches? What is the velocity of the ball when it hits the ground (height 0 )?

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  1. Differentiate to find the max height.

    Velocity is 64 - 32t.

    Max is at s'(t) = 0, when velocity is zero
    s'(t)= 64 - 32t = 0

    so it peaks at t=2. From that you can get the height.

    You now need to find s(t) = 0, so solve 64+64t–16t2 = 0

    That will give you the time, and from that the velocity function 64 - 32t will give you the answer to the second part

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