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The kinematics equation is used to study the motion of an object that moves with a uniform acceleration during its entire motion. If the initial velocity of an object of mass mis u, its final velocity is v, and the object covers a distance s with the uniform acceleration a during its motion, the kinematics equation of motion connecting these variables can be written as

$v^2-u^2=2as.$

The distance between the window of the fourth story of the building and the top of the net is s = 18.0 m.

The distance through which the net is stretched is $螖x=1.0\mathrm{m}$.

The initial velocity of the person is $u=0\mathrm{m}/\mathrm{s}$.

If the downward direction is taken as positive, the acceleration of the person falling downward will be equal to the acceleration due to gravity, i.e.,$a=g=9.8\mathrm{m}/{\mathrm{s}}^2$.

Consider that the final velocity of the person with which she just touches the top of the net is $v\mathrm{m}/\mathrm{s}$.

The person jumps from the window and reaches just the top of the net after traveling a distance of 18.0 m. Using the third equation of motion,

$\begin{array}{c}{v}^2-u^2=2as\\ {\left(v\right)}^2-{\left(0\mathrm{m}/\mathrm{s}\right)}^2=2\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(18.0\mathrm{m}\right)\\ v^2=352.8{\mathrm{m}}^2/{\mathrm{s}}^2\\ v=18.78\mathrm{m}/\mathrm{s}\end{array}$.

The person reaches the top of the net with the velocity of $\mathbf{18}\mathbf{.}\mathbf{78}\mathbf{m}\mathbf{/}\mathbf{s}$.

When the person touches the top of the net, the net is stretched by a distance of 1.0 m to slow down the person to rest. If you consider the motion of the person during the stretching of the net, the initial velocity of the person just before stretching the net is $u\text{'}=v=18.78\mathrm{m}/\mathrm{s}$.

After stretching the net, the person comes to rest. Therefore, the final velocity of the person is$v\text{'}=0\mathrm{m}/\mathrm{s}$.

Using the equation of motion to find the acceleration of the person during the stretching,

$\begin{array}{c}{\left(v\text{'}\right)}^2-{\left(u\text{'}\right)}^2=2a\text{'}\left(螖x\right)\\ {\left(0\mathrm{m}/\mathrm{s}\right)}^2-{\left(18.78\mathrm{m}/\mathrm{s}\right)}^2=2a\text{'}\left(1.0\mathrm{m}\right)\\ a\text{'}=-\frac{352.8{\mathrm{m}}^2/{\mathrm{s}}^2}{2.0\mathrm{m}}\\ =-176.4\mathrm{m}/{\mathrm{s}}^2\end{array}$

Here, the negative sign shows that the person decelerates during stretching. Thus, the average deceleration experienced by the survivor when she was slowed down to the rest is $176.4\mathrm{m}/{\mathrm{s}}^2$.

If you loosen the net, this will provide a larger displacement and more time to the person in order to come to rest.

Thus, the deceleration experienced by the person will be less, and the landing of the person will be safer.