How to calculate the Force exerted by a falling body?

I have been struggling with problem for a while :

Consider a body of mass $M$ released from a height of $h$ meters above the ground. With what amount of Force will it hit the ground ?

My Attempt :

I assume that air-resistance is non-existent and that $h$ is small enough so that the change in $g$ (Acceleration due to Gravity) can be ignored.

Let the body hit the ground with velocity $v$. As its initial velocity is $0$, with a little bit of calculations, it can be found that $v=\sqrt{2gh}$. So, the linear momentum(p) of the body on striking the ground $=Mv=M\sqrt{2gh}$. But the problem here is that the time for which the body is in contact with the ground is not specifically given. If we assume it to be $\Delta t$, the force exerted by the body comes out to be $\dfrac{\Delta p}{\Delta t}=M\dfrac{\sqrt{2gh}}{t}$.

Is there any way of calculating the Force without including the time of contact?

I know that $|F|=\dfrac{dp}{dt}$, but calculus can not be used here since we are not given $p$ as a function of time.

If you had a very squishy object, it will exert a small force on the ground, whereas if you have a very hard object there will be a large force. From this it's clear that asking for the force is ambiguous, we're going to need to introduce some other variable.

As the question suggests, one thing we can do is include a variable, lets call it $\Delta t$, which tells us the duration of the collision between object and ground. The nicest way to do this is to write down the force equation you had:

$F(t) = \frac{dp}{dt}$

I've included the time on the left hand side to remind us that the force will change as a function of time over the course of the collision.What we can do is integrate this equation from $t=0$, the time of contact, to $t=\Delta t$, the time at which the object comes to rest. Then

$\int_0^{\Delta t} F(t) dt = \int_0^{\Delta t} \frac{dp}{dt} dt = \Delta p$

We can multiply and divide by $\Delta t$ to see that

$\Delta p = \Delta t \left(\frac{1}{\Delta t} \int_0^{\Delta t} F(t) dt \right) = \Delta t \,\, F_{avg,t}$

Presumably this is how you found $F = M \sqrt{2g h} / t$, but I wanted to be clear what $F$ meant in that equation.

As an alternative which doesn't use the time, we can exploit the work energy theorem:

$F(x) = \frac{dW}{dx}$

Again, integrate both sides, this time from $x=0$ to $x=\Delta x$, the total distance over which the collision occurs. This time we find

$\Delta x F_{avg,x} = W$

By the work energy theorem $W = \Delta E$ where $\Delta E$ is the change in energy of the object, so

$F_{avg,x} = \frac{1}{2\Delta x} M v^2$

This gives us a way to write down an average force without reference to the time. The tradeoff is that now we have the distance over which the collision occurs, and we find the force averaged over position rather than over time.

The only force that acts on the body is gravitational force. By the time it hits the ground, the force will keep operating $F=Mg$

You can also see this by calculating the time a body takes to fall from a height $h$: $$h=v_ot + \frac{1}{2}gt^2$$ so you have $v_0=0$ and $t= \sqrt{\frac{2h}{g}}$

Now replace it in your equation a have that $F=Mg$

• To me, this answer doesn't seem adequated to what was asked. The question is about the force exerted by the falling body on the ground at the impact, not the force exerted on that body. – JackI Feb 22 '17 at 6:23
Your calculation are correct if you are assuming completely inelastic collision with ground. Otherwise if it is completely elastic (which is never possible in real problem) then change in momentum will be double. For other cases, where collision is in between completely elastic completely inelastic change in momentum will be somewhere in between $M \sqrt{2gH}$ to $2 M \sqrt{2gH}$.