Answer:
Approximately [tex]1.6\times 10^{3}\; \rm N[/tex].
Explanation:
By the Impulse-Momentum Theorem, the change in this woman's momentum will be equal to the impulse that is applied to her.
The momentum [tex]p[/tex] of an object is equal to the product of its mass [tex]m[/tex] and velocity [tex]v[/tex]. That is: [tex]p = m \cdot v[/tex].
Let [tex]v(\text{before})[/tex] and [tex]v(\text{after})[/tex] represent the velocity of the woman before and after the landing. Let [tex]m[/tex] represent the woman's mass.
Therefore, the change in this woman's momentum would be:
[tex]\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}[/tex].
On the other hand, impulse is equal to force multiplied by the duration of the force. Let [tex]F[/tex] represent the average force on the woman. The impulse on her during the landing would be [tex]F \cdot t[/tex].
Apply the Impulse-Momentum Theorem.
Impulse is equal to the change in momentum:
[tex]F \cdot t = m \cdot (v(\text{after})- v(\text{before}))[/tex].
After landing, the woman comes to a stop. Her velocity would become zero. Therefore, [tex]v(\text{after}) = 0\; \rm m \cdot s^{-1}[/tex].
[tex]\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}[/tex].
