This problem can be solved 'numerically' without resorting to other methods. First we have:
$$
w^2 + g^2 = L
$$
$$
(w + f)^2 + (g + p)^2 = L
$$
where $w$ is the wall, $g$ is the ground, $L$ is the ladder squared, $f$ is the fall rate and $p$ is the pull rate. Note that the units of f and p are metres - time is implicit. Also note that we do not presume that the fall rate is negative - it should come out that way. Equating, canceling common terms and solving the quadratic in $f$ yields:
$$
f = \sqrt{w^2 - 2gp - p^2} - w
$$
Now we can calculate a series of increasingly accurate approximations of the fall rate if the pull rate is $0.4 m/s$. Here is a Python function that does this:
def fallrate(wall, ground, pullrate):
for i in [2**n for n in range(15)]:
print(i * ((wall**2 - 2 * ground * (pullrate/i) - (pullrate/i)**2)**.5 - wall))
input("\nEnter to exit. ")
fallrate(4, 3, 0.4)
Note that if we halve the distance we use to calculate the pull rate we will get a rate for half the distance. We must therefore double the result for the purpose of comparison - hence the i * etc
at the start of the approximation formula. The fall rate seems to converge on $-0.3 m/s$. This approach can be seen as a type of calculus - if we divide the rates on both sides of the approximation formula by $i$ we are in effect turning them into infinitesimals; and it agrees with the result given by Eff worked out using conventional methods.