Question

A single cable supporting an unoccupied construction elevator breaks when the elevator is at rest at the top of a 120-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?

Answer 1

The initial velocity of the elevator is zero. The acceleration produced in the elevator is the acceleration due to gravity. Use the third kinematic equation to determine the speed of the elevator in both the case and use the second kinematic equation to determine the time taken by the elevator to reach the respective heights.

Formula used:
$v_2^2 = v_0^2 + 2gh$,
where, ${v_2}$ is the final velocity of the body, ${v_0}$ is the initial velocity, g is the acceleration due to gravity and h is the height of the fall.
Displacement, $h = {v_0}t + \dfrac{1}{2}g{t^2}$,
where, t is the time.

Complete step by step answer:
Since the elevator is falling from the rest position, its initial velocity of the fall is zero. The acceleration produced in the elevator is the acceleration due to gravity. We can solve the entire question using the kinematic equations in the vertical direction.

(a) Using the third kinematic equation of motion, we have,
$v_2^2 = v_0^2 + 2gh$
Here, ${v_2}$ is the final velocity of the elevator when it strikes the ground, ${v_0}$ is the initial velocity, g is the acceleration due to gravity and h is the height of the fall.
Substituting ${v_0} = 0\,{\text{m/s}}$, $g = 9.8\,{\text{m/}}{{\text{s}}^2}$ and $h = 120\,{\text{m}}$ in the above equation, we get,
$v_2^2 = \left( 0 \right) + 2\left( {9.8} \right)\left( {120} \right)$
$\Rightarrow v_2^2 = 2352$
$\therefore {v_2} = 48.5\,{\text{m/s}}$

Therefore, the elevator strikes the ground with speed 48.5 m/s.

(b) We can calculate the time taken by the elevator to strike the ground using the second kinematic equation as,
$h = {v_0}t + \dfrac{1}{2}g{t^2}$
$\Rightarrow h = \dfrac{1}{2}g{t^2}$
$\Rightarrow t = \sqrt {\dfrac{{2h}}{g}}$
Substituting $h = 120\,{\text{m}}$ and $g = 9.8\,{\text{m/}}{{\text{s}}^2}$ in the above equation, we get,
$t = \sqrt {\dfrac{{2\left( {120} \right)}}{{9.8}}}$
$\therefore t = 4.95\,{\text{s}}$

Thus, the elevator was falling for 4.95 seconds.

(c) We can calculate the speed of the elevator at the halfway of the fall using the third kinematic equation as,
$v_1^2 = v_0^2 + 2g\left( {\dfrac{h}{2}} \right)$
$\Rightarrow v_1^2 = 2g\left( {\dfrac{h}{2}} \right)$
Here, ${v_1}$ is the velocity at the halfway of the fall.
Substituting $h = 120\,{\text{m}}$ and $g = 9.8\,{\text{m/}}{{\text{s}}^2}$ in the above equation, we get,
$v_1^2 = 2\left( {9.8} \right)\left( {\dfrac{{120}}{2}} \right)$
$\Rightarrow v_1^2 = 1176$
$\therefore {v_1} = 34.3\,{\text{m/s}}$

Thus, the speed of the elevator at the halfway of the fall is 34.3 m/s.

(d) We can calculate the time to reach the halfway point using the second kinematic equation as,
${t_1} = \sqrt {\dfrac{{2\left( {h/2} \right)}}{g}}$
Substituting $h = 120\,{\text{m}}$ and $g = 9.8\,{\text{m/}}{{\text{s}}^2}$ in the above equation, we get,
${t_1} = \sqrt {\dfrac{{2\left( {120/2} \right)}}{{9.8}}}$
$\therefore {t_1} = 3.50\,{\text{sec}}$

Thus, the time taken by the elevator to reach the halfway point is 3.50 seconds.

Note: In the solution, we have considered the downward motion of the elevator to be the positive direction. One can also solve this question using the law of conservation of energy as, $\dfrac{1}{2}mv_0^2 + mgh = \dfrac{1}{2}mv_2^2 + mg\left( 0 \right)$. In fact, we can derive the kinematic equation from the law of conservation of energy.