- #1

CellCoree

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#1: At what two times is the baseball at a height of 10.9s above the point at which it left the bat? Give your answers in ascending order separated with comma.

my work:

x(t) = X(0) + V(0)t + 1/2at^2

10.9 = 0 + 31.4m/s(t) + 1/2(-9.8)t^2

=-4.9t^2 + 31.4t - 10.9

Using both the Quad. Equation and my calculator, i found the same answer. the answer is: [tex]0.36830144 , 6.0398618[/tex]

can someone help me? i don't know what I am doing wrong