Tug-of-War!

If the height of the pulley is 8 feet and you are standing 15 feet away from the pulley, then by the right triangle formed, 17 feet of the rope is on your side of the pulley, leaving only 10 feet on my side.

Setting up, differentiating, and plugging into the equation that represents the triangle on my side yields $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \longrightarrow 2(6)(5) + 2(8)(0) = 2(10)\left(\frac{dz}{dt}\right) \longrightarrow \frac{dz}{dt} = 3$

Now, the amount of rope being pulled into the hypotenuse of my triangle is equal (but opposite) to the amount of rope being pulled from the hypotenuse of your triangle. So we set up another equation representing the triangle, but this time on your side. We'll then differentiate it and plug in what we know: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \longrightarrow 2(15)\left(\frac{dx}{dt}\right) + 2(8)(0) = 2(17)(3) \longrightarrow \frac{dx}{dt} = 3.4$

This means you're coming closer to me at a rate of 3.4 feet per second.