The figure from below shows a set of three pulleys joined by a cable with one end tied to the roof along one of the pulleys. From the center of the first pulley from the right a block of mass "$m$" is hanging as indicated. Meanwhile the other pulley from the right is supported by a stand where a Force $F$ is pulling to the right. It is known that the mass of the block is $2\,kg$ and it takes $3\,s$ to impact to the ground starting from rest. Given these conditions: Find the value of the force $\vec{F}$. (Assume: $g=10\,\frac{m}{s^2}$).

The alternatives given are as follows:

$\begin{array}{ll}

1.&4\,N\\

2.&6\,N\\

3.&8\,N\\

4.&10\,N\\

5.&12\,N\\

\end{array}$

What i've attempted to do here was to establish this equation.

$F-mg=ma$

But the problem lies on exactly how to define the acceleration in this system. If I go on this route:

$v_f^2=v_o^2+2a\Delta y$

Then

$v_f^2=0+2a\Delta y$

It can be seen from the graph that $\Delta y = 9\,m$

Therefore this can be plug into the previous equation:

$v_f^2=0+2a (9)$

But it also mentions that the time elapsed to hit the ground is $3\,s$. Therefore:

$v_f=v_o+at$

$v_f=0+3a$

If I were to use this into the preceding equation I end up with:

$(3a)^2=0+2a(9)$

And this does produce a contradiction. What am I doing wrong?. Can someone help me here?.