SOLUTION: A ball is thrown into the air vertically with a velocity of 112 feet per second. The ball was released 6 feet above the ground. The height above the ground t seconds after release
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Question 712844: A ball is thrown into the air vertically with a velocity of 112 feet per second. The ball was released 6 feet above the ground. The height above the ground t seconds after release is modeled by h(t)=-16t^2+112t+6. 1.)When will the ball reach 130 feet? 2.)Will the ball reach 250 feet? 3.)In how many seconds after its release will the ball hit the ground? Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! h(t)=-16t^2+112t+6. 1.)When will the ball reach 130 feet?
h(t) = -16t^2 + 112t + 6
-16t^2 + 112t + 6 = 130
Solve for t
(There will be 2 solutions, 130 ft going up, 130 ft coming back down)
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2.)Will the ball reach 250 feet?
-16t^2 + 112t + 6 = 250
Same as #1
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3.)In how many seconds after its release will the ball hit the ground?
-16t^2 + 112t + 6 = 0
Ignore the negative solution
