What is the maximum height from which an object can be dropped without breaking it?

Suppose we drop from rest, a body of mass $$m$$ and breaking stress $$\sigma$$ on the ground from height $$h$$ and it collides with the perfectly rigid ground and comes to rest within a short time $$t$$. The coefficient of restitution is taken to be $$0$$ for simplicity.

For maximum height, we can equate impulsive force to breaking force$$mv/t=\sigma A\hspace{0.3cm}$$ where $$v=\sqrt{2gh}$$

Therefore, $$h=\frac{\sigma^2 A^2 t^2}{2m^2g}$$

Ideally, the body comes to rest momentarily, so $$t=0$$, and maximum height will be $$0$$ for all bodies which doesn't fit with observations. If the object is a human body jumping on his legs from a finite height, his bones don't break even if the collision is almost visibly non-elastic and $$t$$ is apparently too small. What is the reason behind this observation and apparent contradiction with the theory? Is it because of the finiteness of coefficient of restituition or $$t$$ or other simplified assumptions?

How accurate should my model be to give a fairly accurate measure of the maximum height from which an object can be dropped without breaking it?

• Why do you have $t=0$? You can't divide by zero in the first equation. Also, this should be an inequality relationship since $\sigma A$ is a limiting quantity. Nov 20 '20 at 16:30
• Rigid bodies and "perfectly rigid ground" do no exist in real life. To get a sensible answer, you must include elasticity effects (which is not simple). If you want to do an approximate calculation, a simple approximation for an impact between two "hard" bodies (e.g. a solid metal object hitting a concrete floor) would assume the deceleration during the impact was about 30,000 m/s^2 (i.e. 3,000g). Nov 20 '20 at 16:52
• @alephzero Can you please cite a reference for the mentioned not so simple elastic effects? Nov 20 '20 at 18:01

The average impact force is

$$F_{average}=ma_{average}=m\frac{\Delta v}{\Delta t}$$

Clearly, if $$\Delta t=0$$ the average impact force would be infinite.

If the object is a human body jumping on his legs from a finite height, his bones don't break even if the collision is almost visibly non-elastic and $$t$$ is apparently too small. What is the reason behind this observation and apparent contradiction with the theory?

The average impact force can also be related to the stopping distance $$d$$ using the work energy theorem, where $$v$$ is the change in velocity or $$\sqrt {2gh}$$

$$F_{average}d=-\frac{1}{2}mv^2$$

For a given falling height, the human being can reduce his/her average impact force by extending his/her stopping distance during the collision with the ground. An example is by bending the knees after during the collision with the ground. Again, the lower $$d$$ in this equation or $$t$$ in the first equation the greater the average impact force.

We can also consider a rectangular brick instead of a human...When dropped from below a certain height ℎ it doesn't break. I am actually looking for a theoretical approach to determine that minimum height from which when dropped the brick doesn't break

Although you may know the maximum impact force per unit area that the object can withstand, the impact equations can only give you the average impact force. Since the maximum impact force will be greater than the average, you would probably have to reduce the average impact force by some "safety factor" or resort to performing tests

Other than that, to apply the equations you need to know either the duration of the collision, or the stopping distance of the collision.

Hope this helps.

• Nice answer but I dont understand which distance is the stopping distance d? And assuming the human doesn't do any bending or other complicated things to protect himself from a fatal impact how is the maximum height related to the breaking stress of the bones? Nov 20 '20 at 18:17
• We can also consider a rectangular brick instead of a human...When dropped from below a certain height $h$ it doesn't break. I am actually looking for a theoretical approach to determine that minimum height from which when dropped the brick doesn't break Nov 20 '20 at 18:19
• @ManasDogra I updated my answer to respond to your comment. Nov 20 '20 at 18:57

Your equation should be $$h=s^2A^2t^2/2gm^2$$

Your observations about ideal and human bodies do not contradict your theory. By definition ideal bodies are not real, so there are no observations of their behaviour. It is not contradictory for ideal bodies which do not deform (hence $$t=0$$) to break when $$h=0$$. For real bodies if $$t$$ is almost zero then the other values in the equation could nevertheless be large enough to make $$h$$ finite.

To find out how accurate your model is, you need to compare its predictions with the results of an experiment.

It is likely that your model will provide only an order of magnitude estimate. Even if the model is perfectly accurate, predictions could differ from observation because your measurements are very inaccurate. In particular, the value of $$t$$ could be very difficult to measure and could be out by a factor of $$10$$. The prediction of $$h$$ could then be out by a factor of $$100$$.

Your model is based on the average force during the collision. However the object breaks if the maximum force is equal to $$\sigma A$$. The maximum force could be around twice the average, possibly more. Because of this your model gives estimates of height which likely to be 40% higher than observations. If you relied on the prediction your model gives for maximum safe dropping height, the object would probably break.

• How will a more accurate model look like? Can you add something about that or give a reference when you get time? Nov 20 '20 at 18:03
• @ManasDogra Human/animal and inanimate objects behave differently when landing, so you should first decide which kind of object you are trying to model. It will be very difficult to make a model which works well for both. Nov 20 '20 at 18:52
• Oh! I didn't know that. Anyway, I was trying to model inanimate objects originally. Nov 21 '20 at 4:38
• A more accurate model is a mass on a spring colliding with the ground. The spring constant is related to Young's Modulus for the material. You calculate how the force in the spring changes during the collision. This is an elastic collision, but it gives you a start. Nov 21 '20 at 22:53