Free fall

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

Correct answer:

E =  882.9 kJ

Step-by-step explanation:

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

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