If a ball is thrown upward at 64 ft per second from a height of 12 ft, the height of the ball can be modeled by S=12=64t-16^2, where t is the number of seconds after the ball is thrown. how long after the ball is thrown is the height 40 ft?

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v1= 64 feet per second [up]a= 9.81 m/s2 [down]= -9.81 m/s2 [up] = -32.185 feet/s2 [down] (1 metre= 3.28083989501312 feet)t= 3sd= ?Step 2Solve:d= v1t+(1/2)at2d= 64(3)+(1/2)(-32.185)(3)2d= 192-144.8325d= 47.165 feet*Remember you need to add the initial heightd=12+47.165d=59.1675 feet= 10 feet

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