Knowing that during the last second of the fall

Question

A small sphere was abandoned from rest from a height of $ h $ relative to the ground at a location where $ g = 10 m / s ^ 2 $. Knowing that during the last second of the fall, the sphere traveled a distance of $ 35 $ m, determine: a) the speed with which she reached the ground: b) The height $ h $

Does anyone know how to apply Galilean proportions here?

Answer 1

$v(t)=v_0 - gt$ (negative, because down-directed) and $v_0=0$ ("from rest"). And thus (integrating): $h(t)=h_0 - \frac12gt^2$

The fall time is $t=\sqrt{2\frac{h_0}{g}}$, by solving $h(t)=0$. So the height loss in the last second is $h(\sqrt{2\frac{h_0}{g}}-1)$. Set equal to $35$ and solve for $h_0$ and everything can be computed from that.

Answer 2

I had to look up “Galilean proportions” and I am still not quite sure what you mean by them, but I suppose you are referring to the fact that if the sphere falls a distance $y$ in the first second, it falls $3y$ in the second second, $5y$ in the third second, and increasing by $2y$ every additional second.

This would be helpful if we knew the number of seconds that the sphere fell was an integer and we knew how many seconds there were. Since we do not know either of those facts, I do not see a way to apply the proportions directly.

But we do know a few things:

The speed at any instant is $10$ m/s greater than one second earlier.

The distance fallen during each second is $10$ m greater than during the previous second.

The distance fallen in one second (measured in meters) is the arithmetic mean of the speeds (measured in m/s) at the start and end of the one-second period.