Do you feel more impact force when falling straight down or tipping over? Suppose you fall from the top of a ladder straight down.  You will hit the ground with an amount of force.  
Now suppose that you fall over while holding onto the ladder, tipping over in an arc instead of falling straight down.  You will hit the ground with another amount of force.
Neglecting the mass of the ladder and air resistance, which impact will have the most force?  Falling or tipping?
This has been a bit of a debate between myself, my father, and my grandfather.  I believe that they would fall with the same force because, relative to the ground, you start with the same amount of potential energy in both situations.  My grandfather and father, however, guess that it would fall with around half the force because the force is dissipated somewhat by the forward motion and the support of the ladder.  We would appreciate an answer from a third party to help us find the solution.
 A: 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.
A: So the force you would feel is the $F = \frac{\Delta p}{\Delta t}$. Now, when you hit the ground, your momentum basically changes to zero in some sudden time $\Delta t$. Let us assume that the time it takes for the momentum the change to zero ($\Delta t$) is the same for both the cases (tipping over and falling down). Thus $$F = \frac{m \Delta v}{\Delta t} = \frac{m (v-0)}{\Delta t} = \frac{m v }{\Delta t}\;.$$
Now when falling down, all your initial potential energy is converted to KE implies $$ v = \sqrt{2 g h} \implies F  = \frac{ m \sqrt{2 g h} }{ \Delta t}\;.$$
When it tips over, you now have both KE and Rotational energy. We then have $$\frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 = m g h = \frac{1}{2} m v^2 + \frac{1}{2} (m h^2) \left(\frac{v}{h} \right)^2 = m v^2 \implies v = \sqrt{gh}\;.$$
We then have $$ F = \frac{m \sqrt{gh}}{\Delta t}\;.$$
Thus $$\frac{ F_\textrm{fall} } {F_\textrm{tip} } = \sqrt{2} \implies F_\textrm{fall } > F_\textrm{tip}\;.$$
So falling down imparts a greater force!
