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the position of a ball thrown

label Calculus
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schedule 0 Hours
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the position of a ball thrown up from the top of a building 150 ft tall with an initial velocity of 20 ft/sec is given by the function s(t) = -16r^2 +20t +150. the maximum height of the ball is 

Apr 8th, 2015

The height of the ball is described by a quadratic function s(t) = at^2 + bt + c, where a = -16, b = 20, and c =150. The function attains its maximum at t = -b/(2a) = 20/(2*16) = 5/8 = 0.625 s.

Then the height is s(0.625) = - 16*(5/8)^2 + 20*(5/8) + 150 = 156.25 ft above the ground.


Apr 8th, 2015

thanks 

Apr 8th, 2015

you are welcome

Apr 8th, 2015

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Apr 8th, 2015
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