The ball will move following the law of the accelerated linear motion, where the acceleration is given by the force of gravity.
If we draw the vertical axis growing positively upwards, the acceleration will be negative, and the general formula
[tex] s(t) = s_0 + v_0t + \dfrac{1}{2}at^2 [/tex]
becomes
[tex] s(t) = 13.7t-9.8t^2 [/tex]
This is the equation of a parabola, concave down, where the vertex is the maximum. In this case, the vertex is reached when
[tex] t = \dfrac{-13.7}{2\cdot (-9.8)} = \dfrac{13.7}{19.6} \approx 0.7 [/tex]
And if we plug this value for t we have
[tex] s(0.7) = 13.7\cdot 0.7-9.8\cdot 0.7^2 \approx 4.8 [/tex]
So, the ball passes through half the maximum weight (i.e. 2.4) when the following equation is solved:
[tex] 2.4 = 13.7t-9.8t^2 \iff 9.8t^2-13.7t+2.4 = 0 [/tex]
whose solutions are approximately
[tex] t=0.2,\quad t=1.2 [/tex]
