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
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(174) (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.
