Answer
$9.6~m$
Work Step by Step
Let the origin be on the ground with the positive direction upward.
The motion of free falling objects is symmetrical, that is, the time it takes to go from the floor to the top is the same time it takes to the object move back from the top and reach the floor. So, the ball takes $\frac{2.8~s}{2}=1.4~s$ to reach the top (half of the total time).
First, let's find the ball's initial speed:
$v=0$, $a=-g=-9.81~m/s^2$, $t=1.4~s$
$v=v_0+at$
$0=v_0+(-9.81~m/s^2)(1.4~s)$
$v_0=(9.81~m/s^2)(1.4~s)=13.7~m/s$
$v^2=v_0^2+2a\Delta x$
$0^2=(13.7~m/s)^2+2(-9.81~m/s^2)\Delta x$
$\Delta x=\frac{(13.7~m/s)^2}{2(9.81~m/s^2)}=9.6~m$
