Both you and your ancestors are wrong. But I bet you would never guess the real answer!
Assuming the base of the ladder doesn't slide you have a rotating system. Just like a freely falling man you convert potential energy to kinetic energy, but for a rotating system the kinetic energy is given by:
$$ T = \frac{1}{2}I\omega^2 $$
where $I$ is the angular momentum and $\omega$ is the angular velocity. Note that $v = r\omega$, where $r$ is the radius (i.e. the length of the ladder). We'll need this shortly.
Let's start by ignoring the mass of the ladder. In that case the moment of inertia of the system is just due to the man and assuming we treat the man as a point mass $I = ml^2$, where $m$ is the mass of the man and $l$ the length of the ladder. Setting the change in potential energy $mgl$ equal to the kinetic energy we get:
$$ mgl = \frac{1}{2}I\omega^2 = \frac{1}{2}ml^2\omega^2 = \frac{1}{2}mv^2 $$
where we get the last step by noting that $l\omega = v$. So:
$$ v^2 = 2gl $$
This is exactly the same result as we get for the man falling straight down, so you hit the ground with the same speed whether you fall straight down or whether you hold onto the ladder.
But now let's include the mass of the ladder, $m_L$. This adds to the potential energy because the centre of gravity of the ladder falls by $0.5l$, so:
$$ V = mgl + \frac{1}{2}m_Lgl $$
Now lets work out the kinetic energy. Since the man and ladder are rotating at the same angular velocity we get:
$$ T = \frac{1}{2}I\omega^2 + \frac{1}{2}I_L\omega^2 $$
For a rod of mass $m$ and length $l$ the moment of inertia is:
$$ I_L = \frac{1}{3}m_Ll^2 $$
So let's set the potential and kinetic energy equal, as as before we'll substitute for $I$ and $I_L$ and set $\omega = v/l$. We get:
$$ mgl + \frac{1}{2}m_Lgl = \frac{1}{2}mv^2 + \frac{1}{6}m_Lv^2$$
and rearranging this gives:
$$ v^2 = 2gl \frac{m + \frac{1}{2}m_L}{m + \frac{1}{3}m_L} $$
and if $m_L \gt 0$ the top of the fraction is greater than the bottom i.e. the velocity is greater than $2gl$. If you hold onto the ladder you actually hit the ground faster than if you let go!
This seems counterintuitive, but it's because left to itself the ladder would rotate faster than the combined system of you and the ladder. In effect the ladder is accelerating you as you and the ladder fall. That's why the final velocity is higher.
Best Answer
Of course it will. The effective velocity of the hail is the vector sum of the velocity of the falling hail and the moving car - this means the windshield will be hit by hail going faster and thus having more momentum. To reverse this momentum will require a greater $F\Delta t$ .
Note - I did say "hit the windshield" - or more generally any forward facing component. It won't change the force on the roof and there may be areas like the rear window that will experience less impact (maybe none at all).
If your car is a rectangular box, then in the stationary case there is no hail hitting the front or rear; just the top. When the car (box) is moving, the same number of hail stones will be hitting the roof per unit time, and their vertical velocity will be the same. Assuming that the hail stone bounce off with a perfectly elastic collision (no sliding friction), then the force on the roof is unchanged.
The same is not true for the front of the car / box. When it was stationary, nothing hit the front; now that it is moving, the car "intercepts" hail that is falling straight down.
You can calculate the force due to this if you know the density of the hail (mass of hail / unit volume). The volume of air-with-hail that is swept by the car in unit time is
$$vol = v_{car}A_{car}$$
In words: the volume equals the velocity times the area of the car (the front facing surface of the "box"). The total amount of hail that is hit is thus
$$m_{hail}=vol \cdot \rho_{hail}$$
and since the velocity of the hail changes by $2v_{car}$ (elastic collision assumed), the change of momentum in the car's frame of reference is $2 m_{hail} v_{car}$. Since this is in unit time, the (new) force on the car due to hitting the hail is
$$F=2 \rho_{hail} v_{car}^2 A_{car}$$
For any $v_{car}>0$ this says there is an additional force from the hail on the car because it is moving.