SOLUTION: At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maxi

Question 571884: At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maximum height of the ball? And how long will it take to reach its maximum height?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
At 9:45 am Sue threw a ball upwards while standing on a platform 35 feet above the ground.The height after x seconds follows the equation H(x)=-0.6X^2+72x+35. What will be the maximum height of the ball? And how long will it take to reach its maximum height?
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The vertex occurs where x = -b/(2a) = -72/(-1.2) = 6 seconds
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Maximum height = H(6) = 445.4 ft
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Cheers,
Stan H.
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