I work for a company selling tethers that prevent tools on building sites from falling away from the user.

One of the problems that has come up is that objects can deflect horizontally when they strike an object whilst falling. You can imagine that they've fallen and then hit a steel girder or something on the way down.

Given that the weight of the object, the height dropped from and the impact height are known, how would I calculate the possible deflection distance?


1 Answer 1


If we are going for the maximum range, we could treat the collisions with things on the way down as elastic, and in the worst case, the collision would preserve the kinetic energy of the object, but change its direction so that it might move in a different direction.

The first problem to solve is that the range is for a projectile launched from a given height $y_0$, with an initial velocity $v$ and launch angle $\theta$. The answer is:

$$ d = \frac{ v \cos \theta }{g} \left( v \sin \theta + \sqrt{ v^2 \sin^2 \theta + 2 g y_0 } \right) $$

Next, we are just interested in the worst case, so we need to maximize this over $\theta$ by setting its derivative to zero and substituting back in, we obtain:

$$ d = \frac{v}{g} \sqrt{ v^2 + 2 g y_0 } $$ for the maximum range of a projectile launched with velocity $v$ from height $y_0$.

Now, if the original object were falling from some higher initial height $h$, if the collision with the beam or girder were completely elastic, by conservation of energy, we expect a maximum velocity coming off the beam of:

$$ \frac 12 mv^2 = mg (h - y_0) \implies v^2 = 2 g (h-y_0) $$

substituting this back into our maximum range, we have

$$ \boxed{ d = 2 \sqrt{ h (h-y_0) } }$$

Sketched out: Dropped Wrench

Here I've plotted the height the collision $(y_0)$ occurs versus the maximum distance the object is expected to travel in this model $(d)$, both as a percentage of the total height $(h)$. Sketched out, it becomes clear that the furthest we expect the object to travel is twice the total height, and this occurs if it takes a bad bounce once it reaches the ground, at higher collision heights, it can't go as far because it hasn't gained enough speed in its fall yet.

  • $\begingroup$ That makes sense. I saw a graph showing an object falling 200ft and impacting at 20ft then travelling 400ft horizontally. It also indicated that it could bounce 60ft higher in the air after impact, which I found surprising. $\endgroup$
    – jidl
    Sep 6, 2014 at 10:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.