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?

1a) -192.7g

1b) 0.0126 s

2) 1309 kg m/s

3)

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:

where

m = 55.0 kg is the mass of the woman

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,

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

So we can find the deceleration using the suvat equation:

where

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

Solving for a,

In terms of g,

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:

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

is the acceleration of the woman

t is the time of the impact

Solving for t, we find:

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:

where

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:

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

where

I is the impulse

F is the average force on the object

is the time of the collision

Here we have:

is the magnitude of the impulse

is the duration of the collision

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

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air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.