SOLUTION: Baseballs from straight upward with an initial speed of 64 feet a second the number of feet S above the ground after T seconds is given by the equation as equals -16 T squared +64t
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Question 989068: Baseballs from straight upward with an initial speed of 64 feet a second the number of feet S above the ground after T seconds is given by the equation as equals -16 T squared +64t. when will the baseball be 48 feet above ?the ground when will he hit the ground?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Baseballs from straight upward with an initial speed of 64 feet a second the number of feet S above the ground after T seconds is given by the equation as equals -16 T squared +64t. when will the baseball be 48 feet above ?the ground when will he hit the ground?
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h(t) = -16t^2 + 64t
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-16t^2 + 64t = 48
Solve for t
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-16t^2 + 64t = 0
Solve for t
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