# SOLUTION: The height in feet above the ground of a ball thrown straight up with an initial velocity of 96 ft/sec is given by the formula h(t)=-16t^2 + 96t + 4. where T is the time, in secon

Algebra ->  -> SOLUTION: The height in feet above the ground of a ball thrown straight up with an initial velocity of 96 ft/sec is given by the formula h(t)=-16t^2 + 96t + 4. where T is the time, in secon      Log On

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 Click here to see ALL problems on Quadratic Equations Question 134506: The height in feet above the ground of a ball thrown straight up with an initial velocity of 96 ft/sec is given by the formula h(t)=-16t^2 + 96t + 4. where T is the time, in seconds, after the ball is thrown. find the values of T for which the height of the ball is 48 feet. find the exact solutions.Answer by rajagopalan(158)   (Show Source): You can put this solution on YOUR website!h(t)=-16t^2 + 96t + 4. 48=-16t^2+96t+4 -16t^2+96t+4=48 -16t^2+96t+4-48=0 -16t^2+96t-44=0 divide thro by 4 -4t^2+24t-11=0 split mid tem into 22t+2t -4t^2+24t-11=0 -4t^2+22t+2t-11=0 -2t(2t-11)+1(2t-11)=0 (-2t+1)(2t-11)=0 giving (-2t+1)=0 and t=0.5 2t-11=0 giving t=5.5 Ans 0.5 and 5.5 seconds from start of throw respectively for onward and return paths.