SOLUTION: a ball is thrown vertically upward from a height of 6ft. with an initial velocity of 64ft/s, its height h after t seconds is given by h = -16t^2 + 64t + 6. How long does it take t

Question 184916: a ball is thrown vertically upward from a height of 6ft. with an initial velocity of 64ft/s, its height h after t seconds is given by h = -16t^2 + 64t + 6. How long does it take the ball to reach a height of 38 ft. on the way up?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a ball is thrown vertically upward from a height of 6ft. with an initial velocity of 64ft/s, its height h after t seconds is given by h = -16t^2 + 64t + 6. How long does it take the ball to reach a height of 38 ft. on the way up?
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h(t) = -16t^2 + 64t + 6
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Let the height be 38 and solve for "t":
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-16t^2 + 64t + 6 = 38
-16t^2 + 64t - 32 = 0
Divide thru by -16 to get:
t^2 - 4t +2 = 0
t = [4 +- sqrt(16 -4*1*2)}/2
t = [4 +- sqrt(8)]/2
Solve for the smaller of the two t values:
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t = [4 - sqrt(8)]/2 = 0.586.. seconds
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Cheers,
Stan H.

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