Question

There are many well-documented cases of people surviving falls from heights greater than 20.0 m. In one such case, a 55.0 kg woman survived a fall from a 10th floor balcony, 29.0 m above the ground, onto the garden below, where the soil had been turned in preparation for planting. Because of the "give" in the soil, which the woman compressed a distance of 15.0 cm upon impact, she survived the fall and was only briefly hospitalized.
1. What was the magnitude of her deceleration during the impact in terms of g? Assuming a constant acceleration, what was the time interval (in s) during which the soil brought her to a stop?
2. What was the magnitude of the impulse (in N · s) felt by the woman during impact?
3. What was the magnitude of the average force (in N) felt by the woman during impact?

Answer 1

1a) -192.7g

1b) 0.0126 s

2) 1309 kg m/s

3) $1.04\cdot 10^5 N$

Explanation:

1a)

First of all, we have to find the velocity of the womena just before hitting the ground.

Since the total mechanical energy is conserved during the fall, the initial gravitational potential energy of the woman when she is at the top is entirely converted into kinetic energy.

So we can write:

$mgh=\frac{1}{2}mv^2$

where

m = 55.0 kg is the mass of the woman

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 29.0 m is the initial height of the woman

v is her final speed

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(29.0)}=23.8 m/s$

Then, when the woman hits the soil, she is decelerated until a final velocity

$v'=0$

So we can find the deceleration using the suvat equation:

$v'^2-v^2=2as$

where

s = 15.0 cm = 0.15 m is the displacement during the deceleration

Solving for a,

$a=\frac{v'^2-v^2}{2s}=\frac{0-23.8^2}{2(0.15)}=-1888.3 m/s^2$

In terms of g,

$a=\frac{-1888.3}{9.8}=-192.7g$

1b)

Here we want to find the time it takes for the woman to stop.

Since her motion is a uniformly accelerated motion, we can do it by using the following suvat equation:

$v'=v+at$

where here we have:

v' = 0 is the final velocity of the woman

v = 23.8 m/s is her initial velocity before the impact

$a=-1888.3 m/s^2$ is the acceleration of the woman

t is the time of the impact

Solving for t, we find:

$t=\frac{v'-v}{a}=\frac{0-23.8}{-1888.3}=0.0126 s$

So, the woman took 0.0126 s to stop.

2)

The impulse exerted on an object is equal to the change in momentum experienced by the object.

Therefore, it is given by:

$I=\Delta p =m(v'-v)$

where

$\Delta p$ is the change in momentum

m is the mass of the object

v is the initial velocity

v' is the final velocity

Here we have:

m = 55.0 kg is the mass of the woman

v = 23.8 m/s is her initial velocity before the impact

v' = 0 is her final velocity

So, the impulse is:

$I=(55.0)(0-23.8)=-1309 kg m/s$

where the negative sign indicates the direction opposite to the motion; so the magnitude is 1309 kg m/s.

3)

The impulse exerted on an object is related to the force applied on the object by the equation

$I=F\Delta t$

where

I is the impulse

F is the average force on the object

$\Delta t$ is the time of the collision

Here we have:

$I=1309 kg m/s$ is the magnitude of the impulse

$\Delta t = 0.0126 s$ is the duration of the collision

Solving for F, we find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{1309}{0.0126}=1.04\cdot 10^5 N$