Question 409756
If a ball is thrown straight up into the air with an initial velocity of 60 ft/s, its height in feet after t seconds is given by
y=60t−16t2 
When is the ball at its maximum height?
Max occurs when x = -b/(2a) = -60/(2*-16) = 60/32 = 15/8 seconds
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After 15/8 = 1.875 seconds.
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What is the maximum height attained by the ball?
f(1.875) = 60(1.875)-16(1.875)^2 = 56.25 ft.
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Cheers,
Stan H.
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