SOLUTION: Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.) A ball is thrown vertically upward from a height of 5 feet with an initial velocity of 46 fe

Question 286864: Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 5 feet with an initial velocity of 46 feet per second. How high will the ball go? (Round your answer to two decimal places.)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Use a(t) = -32 ft/sec2 as the acceleration due to gravity. (Neglect air resistance.)
A ball is thrown vertically upward from a height of 5 feet with an initial velocity of 46 feet per second. How high will the ball go? (Round your answer to two decimal places.)
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h(t) = -16t^2 + 46t + 5
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Top of the arc is the vertex.
Vertex occurs when x = -b/2a = -46/(-32) = 23/16 seconds.
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Height at the vertex = h(23/16) = -16(23/16)^2+46(23/16)+5 = 38.06 ft.
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Cheers,
Stan H.
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