An object of mass m is dropped from a tower of height d. Show that according to
quantum mechanics the closest the object can fall on average to the base of the
I don't know - that is the problem to know what to use.
The Attempt at a Solution
Using v^2=u^2+ 2as => velocity when mass reaches ground is v=√2gd
so momentum p=mv=m√2gd
using Uncertainty Principle ∆x∆p=h => ∆x=h/∆p=h/(m√2gd)
This is not the right answer so what have I done wrong?