SOLUTION: A ball is thrown upward with an initial velocity of 32 ft/s from a height 768 ft. h=-16t^2+32t+768. After how many seconds will the ball reach the ground?
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Question 265826: A ball is thrown upward with an initial velocity of 32 ft/s from a height 768 ft. h=-16t^2+32t+768. After how many seconds will the ball reach the ground? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A ball is thrown upward with an initial velocity of 32 ft/s from a height 768 ft. h=-16t^2+32t+768.
After how many seconds will the ball reach the ground?
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The height is zero when the ball hits the ground.
Solve -16t^2 + 32t + 768 = 0 for a positive "t" value.
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-16(t^2 - 2t - 48) = 0
t^2 -2t - 48 = 0
(t-8)(t+6) = 0
Positive solution:
t = 8 seconds
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Cheers,
Stan H.
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