Height of a baseball - kinematics

[CellCoree]

A major leaguer hits a baseball so that it leaves the bat at a speed of 31.4 m/s and at an angle of 38.1 above the horizontal. You can ignore air resistance

#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: $$0.36830144 , 6.0398618$$

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

 

[Leong]

Supposed to be 10.9 m in your question.
The velocity of 31.4 m/s is the vector sum of the x and y components. in your equation, you should use the y component of the velocity which is given as 31.4*sin 38.1.