Free fall

The elevator car weighing 720 kg is torn off at a height of 125 m and begins to fall in a free fall. How much kinetic energy does the elevator car have when it hits the ground floor of the building, when the free fall of the car lasted 5 s?

Correct result:

E =  882.9 kJ

Solution:

$$\begin{gathered} m=720\text{ kg } \\ h=125\text{ m } \\ g=9.81{\text{m/s}}^2 \\ h={\displaystyle \frac{1}{2}}g{t}^2 \\ t=\sqrt{2\cdot h\mathrm{/}g}=\sqrt{2\cdot 125\mathrm{/}9.81}\doteq 5.0482\text{ s } \\ v=g\cdot t=9.81\cdot 5.0482\doteq 49.5227\text{ m/s } \\ {E}_1={\displaystyle \frac{1}{2}}\cdot m\cdot v^2={\displaystyle \frac{1}{2}}\cdot 720\cdot 49.5227^2\doteq 882900\text{ J } \\ E=E_1\to \text{ kJ}=E_1\mathrm{/}1000\text{ kJ}=882900\mathrm{/}1000\text{ kJ}=882.9\text{ kJ}={\displaystyle \frac{8829}{10}}\text{ kJ} \end{gathered}$$

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