Question

A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.

Answer 1

Answer:

Approximately $1.6\times 10^{3}\; \rm N$.

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 $p$ of an object is equal to the product of its mass $m$ and velocity $v$. That is: $p = m \cdot v$.

Let $v(\text{before})$ and $v(\text{after})$ represent the velocity of the woman before and after the landing. Let $m$ represent the woman's mass.

• The woman's momentum before the landing would be $m \cdot v(\text{before})$.
• The woman's momentum after the landing would be $m \cdot v(\text{after})$.

Therefore, the change in this woman's momentum would be:

$\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}$.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let $F$ represent the average force on the woman. The impulse on her during the landing would be $F \cdot t$.

Apply the Impulse-Momentum Theorem.

• Impulse: $F\cdot t$.
• Change in momentum: $m \cdot (v(\text{after})- v(\text{before}))$.

Impulse is equal to the change in momentum:

$F \cdot t = m \cdot (v(\text{after})- v(\text{before}))$.

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, $v(\text{after}) = 0\; \rm m \cdot s^{-1}$.

$\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}$.