Question

A cement block accidentally falls from rest from the ledge of a 53.4-m-high building. When the block is 14.2 m above the ground, a man, 1.91 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?

Answer 1

DATA:

$v_{0}=0$

$t=?$

Solution:

see the following figure:

First, we calculate the distance $y_{1}$ the block since released until the man sees it

$y_{1}=y-y_{2}$

Now, we calculate the speed of the block $v_{1}$ when it has fallen 39.2 m:

$v_{1}^{2}=v_{0}^{2}+2.g.y_{1}$

$v_{1}=\sqrt{v_{0}^{2}+2.g.y_{1}}$

Replacing values given in data:

$v_{1}=\sqrt{(0)^{2}+2.(9.81\, m/s^{2}).(39.2\, m)}$

Second, we calculate the distance $y_{3}$ down the block to the man's head:

$y_{3}=y_{2}-y_{m}$

Now, we calculate the speed of the block $v_{2}$ when it has fallen a distance $y_{3}$ :

${v_{2}}^{2}={v_{1}}^{2}+2.g.y_{3}$

$v_{2}=\sqrt{{v_{1}}^{2}+2.g.y_{3}}$

Replacing values given in data:

$v_{2}=\sqrt{(27.7\, m/s)^{2}+2.(9.81\, m/s^{2}).(12.3\, m)}$

Finally, the time (at most) does the man have to get out of the way is calculated using:

$v_{2}=v_{1}+g.t$

Isolating :

$t=\frac{v_{2}-v_{1}}{g}$

Replacing values:

$t=\frac{31.8\, m/s-27.7\, m/s }{ 9.81\, m/s^{2}}$

${\color{Blue} t=0.418\: s}$