Answer
(a) More than $2.0~m$.
(b) $8.0~m$
Work Step by Step
(a) We both are under the same acceleration: $g$. The increase of our speed is given by: $\Delta v=at=gt$. So, the speed of my friend relative to me remains the same during the fall until she hits the water. Consequently, my friend moves away from me every moment.
(b) First, let's find the total height of the fall.
Let the positive direction be downward and the origin be the bridge.
$x_0=0$, $v_0=0$, $t=1.6~s$, $a=g=9.81~m/s^2$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x=0+0t+\frac{1}{2}(9.81~m/s^2)(1.6~s)^2=12.56~m$
Now, let's find the separation between me and my friend. In this last part, the initial instant is when I step off the bridge.
$x=x_0+v_0t+\frac{1}{2}at^2$
$2.0~m=0+0t+\frac{1}{2}(9.81~m/s^2)t^2$
$t=\sqrt {\frac{2(2.0~m)}{9.81~m/s^2}}=0.639~s$. It means I jumped $0.639~s$ after my friend's jump. So, we are going to fall together during: $1.6~s-0.639~=0.961~s$
$x=x_0+v_0t+\frac{1}{2}at^2$.
$x_{me}=0+0(0.961~s)+\frac{1}{2}(9.81~m/s^2)(0.961~s)^2=4.53~m$. At this moment, my friend reaches the water. That is:
$x_{friend}=12.56~m$
Finally:
$x_{friend}-x_{me}=8.0~m$
