Question

A stone is dropped down a well and hits the water 2.50 s later. What is the depth from the edge of the well to the water?

Answer 1

We know that all objects accelerate at the constant rate in the Earth's gravitational field. This acceleration is denoted with g and is equal to:

`g=9.81(m)/(s^2)`
It seems illogical but the feather and the stone would actually hit the ground at the same time if dropped from the same height (in the vacuum of course).
We also know that distance traveled by the object under constant acceleration is:

`S=(at^2)/(2)`
In our case the object is falling down in the gravitational field with constant acceleration g:

`H=g(t^2)/(2)=9.81(2.5^2)/(2)=30.65$m`