91Ó°ÊÓ

The height of the window ledge above the floor is h = 5 m

The mass of stuntmen is: M = 80 kg

The mass of villains is: m = 70 kg

The coefficient of kinetic friction between bodies and the floor is ${\mathrm{μ}}_k=0.250$.

The momentum conservation gives the speed of entwined foes and the work-energy theorem gives sliding distance on the floor.

The speed of the stuntman is given as:

$u=\sqrt{2gh}$

Here, g is the gravitational acceleration.

Substitute all the values in the above equation.

$$\begin{gathered} u=\sqrt{2\left(9.8m/s^2\right)\left(5m\right)} \\ u=9.9m/s \end{gathered}$$

Apply the momentum conservation to find the speed of entwined woes.

$Mu=\left(m+M\right)v$

Substitute all the values in the above equation.

$$\begin{gathered} \left(80kg\right)\left(9.9m/s\right)=\left(80kg+70kg\right)v \\ v=5.3m/s \end{gathered}$$

Therefore, the speed of entwined woes on floor is 5.3 m/s .

Apply the work energy theorem to find slide distance.

$$\begin{gathered} \frac{1}{2}\left(m+M\right)v^2={\mathrm{μ}}_k\left(m+M\right)gd \\ d=\frac{v^2}{2{\mathrm{μ}}_{k}g} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} d=\frac{{\left(5.3m/s\right)}^2}{2\left(0.250\right)\left(9.8m/s^2\right)} \\ d=5.73m \end{gathered}$$

Therefore, the sliding distance for entwined woes is 5.7 m .