SOLUTION: Baseballs from straight upward with an initial speed of 64 feet a second the number of feet S above the ground after T seconds is given by the equation s= -16 T squared +64t. when
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Question 989069: Baseballs from straight upward with an initial speed of 64 feet a second the number of feet S above the ground after T seconds is given by the equation s= -16 T squared +64t. when will the baseball be 48 feet above ?the ground when will he hit the ground? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! h=-16t^2+64t
h=48
Therefore, -16t^2+64t=48,
16t^2-64t+48=0
dividing by 16
t^2-4t+3=0
(t-3)(t-1)=0
and t=1 and 3 seconds it will be 48 feet above the ground.
It will hit the ground when -16t^2+64t=0, or t=4 seconds.
