Question: The single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a $\mathbf{120}\mathbf{}\mathbf{m}$ high building. (a) With what speed does the elevator strike the ground? (b) How long is it falling? (c) What is its speed when it passes the halfway point on the way down? (d) How long has it been falling when it passes the halfway point?

$\mathrm{y}=120\mathrm{m}$

Initial velocity$\left({\mathrm{v}}_0\right)=0$

The problem deals with the kinematic equations of motion. Kinematics is the study of how a system of bodies moves without taking into account the forces or potential fields that influence the motion. The equations which are used in the study are known as kinematic equations of motion.

Formula:

The final velocity is given by,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{ay}.............\left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}.............\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{ay}................\left(\mathrm{i}\right) \\ {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at}................\left(\mathrm{ii}\right) \end{gathered}$$

To find the speed of the elevator when it strikes the ground, use the following formula

Using equation (i)

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2=0+2\times 9.81\times 120 \\ {\mathrm{v}}_{\mathrm{f}}^2=48.5\mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity when it reaches the bottom is $48.5\mathrm{m}/\mathrm{s}$.

To find time of elevator when it strikes ground, use the following formula

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{a}\times \mathrm{t} \\ \Rightarrow 48.5=0+9.81\times \mathrm{t} \\ \Rightarrow \mathrm{t}=4.95\mathrm{s} \end{gathered}$$

Time to reach the bottom is $4.95\mathrm{s}$.

To find the speed of the elevator when it is halfway (y = 60m) , use the following formula

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\times \mathrm{a}\times \mathrm{y} \\ \Rightarrow {\mathrm{v}}_{\mathrm{f}}^2=0+2\times 9.80\times 60 \\ \Rightarrow {\mathrm{v}}_{\mathrm{f}}=34.3\mathrm{m}/\mathrm{s} \end{gathered}$$

Velocity at the half-way mark is $34.3\mathrm{m}/\mathrm{s}$.

To find the time of the elevator when it is halfway, use the following formula

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_0+\mathrm{at} \\ \Rightarrow 34.3=0+9.81\times \mathrm{t} \\ \Rightarrow \mathrm{t}=3.50\mathrm{s} \end{gathered}$$

Time to reach the half way mark is$3.50\mathrm{s}$