# 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

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 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(57962)   (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.) ------------------ h(t) = -16t^2 + 46t + 5 --- Top of the arc is the vertex. Vertex occurs when x = -b/2a = -46/(-32) = 23/16 seconds. ---- Height at the vertex = h(23/16) = -16(23/16)^2+46(23/16)+5 = 38.06 ft. ======================= Cheers, Stan H.