Question

A man steps and falls freely from rest from a tower to the ground. If he falls half the total height of the tower in last 1 second of his motion. Find height h of the tower?

Answer 1

Answer on Question #47750, Physics, Mechanics — Kinematics — Dynamics

A man steps and falls freely from rest from a tower to the ground. If he falls half the total height of the tower in last 1 second of his motion. Find height h of the tower?

Solution

Let us relate velocity that he had exactly before 1 s of falling and height h of the tower. We have

$h/2=v_{b}\cdot 1\,s+g(1\,s)^{2}/2$

But from other hand, this velocity was obtained in first $t_{b}$ seconds, hence

$v_{b}=gt_{b}$

and during $t_{b}$ he passed also $h/2$:

$h/2=gt_{b}^{2}/2$

So we have

$t_{b}=\sqrt{\frac{h}{g}}$

$v_{b}=g\sqrt{\frac{h}{g}}=\sqrt{hg}$

And combining with first equation:

$h=\sqrt{hg}/2\cdot(1\,s)+9.8\cdot\cdot(1\,s)^{2}/2$

$h^{2}=hg/4+(9.8/2)^{2}$

Solving this equation with respect to h will yield:

$h\approx 6.3\,m$