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anonymous

  • 4 years ago

If a ball is thrown vertically upward from the roof of 32 foot building with a velocity of 96 ft/sec, its height after t seconds is s(t)=32+96t−16t^2. a) What is the maximum height the ball reaches? b) What is the velocity of the ball when it hits the ground (height 0)?

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  1. anonymous
    • 4 years ago
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    The ball reaches its maximum height when it changes direction from going up to going downwards. At that point the velocity is equal to zero so you need to find the velocity equation and equal it to zero by finding the derivative. v(t)= s'(t) = 96-32t=O Solve for t. t is the time when it reaches the highest point. Now substitute this time to the position function to find the height.

  2. anonymous
    • 4 years ago
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    what's t? is it 96 or 32?

  3. anonymous
    • 4 years ago
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    96-32t=0 -32t=-96 t=3

  4. anonymous
    • 4 years ago
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    plug this value into s(t)

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