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.
You probably know the equation for the drag, but just for record it's:
$$ F_{drag} = \frac{1}{2} \rho \space C_d A \space v^2 $$
and rearranging this gives:
$$ \frac{2F_{drag}}{\rho \space C_d \space v^2} = A $$
and you're given $C_d$ (1.4) and $v$ (3 m/s). I would guess you're meant to take the density of Earth's atmosphere at sea level, and a quick Google gives this as 1.2754 kg/m$^3$ (is the density of the Martian atmosphere really 2/3 that of Earth? I thought it was more like 1%). The force is just the weight of the probe (40kg) multiplied by the acceleration due to gravity at the surface of Mars ($g_{Mars} = 3.75\ m/s^2$). So you have everything you need to calculate $A$.
It's not obvious to me why you need the speed of the probe without the parachute ...
Best Answer
The answer to this is relatively simple. Assuming an inexisting atmosphere (and thus terminal velocity) on mars, the kinetic energy of the weight equals its potential energy.
Potential energy is calculated relatively simple:
$$E_p = m \cdot g \cdot h$$
now if you wanted to find out how fast the mass would impact on mars, you can set this equal to the kinetic energy you get after completely using up that potential (aka. when you fall 100 m):
$$ \frac{1}{2} \cdot m \cdot v^2 = m \cdot g \cdot h$$
now with a little standard mathematics, you arrive at a formula for your velocity, depending on height of where you start from:
$$ v = \sqrt{2 \cdot g \cdot h} $$
Interestingly (and as mentioned in the already existing answer) the velocity can be removed from the equation. For the sake of the demonstration let's calculate it either way:
$$ v = \sqrt{2 \cdot 3.711 \frac{m}{s^2} \cdot 100 m} = \sqrt {742.2 \frac {m^2}{s^2}} = 27.24... \frac {m}{s}$$
This is a roundabout $100 \frac{km}{h}$, give or take. (or a little more than 60 mph).
Anyways since you wanted to know about the "Force" of the collision...
How much force does it take to stop $100kg$ with a velocity of $100 \frac{km}{h}$?
The question here is... How fast do you want to stop? The slower you stop, the less continuously applied force is required.
When we multiply our velocity and the mass we get an Impulse (which is $F \cdot\Delta t$ or $m \cdot v $). We now know the impulse of our mass: $2700 Ns$
To get this stopped in a tenth of a second (which is an extremely optimistic guess, assuming falling on the hard mars), you need a whopping: $27 kN$