The force of a Hurricane on a person

[kudoushinichi88]

Homework Statement

During a hurricane, winds can whip horizontally at speeds of 150 km/h. Given that
the density of air is 1.20 kg/m3 and the wind after striking the person is brought to
rest, calculate the force of the wind on the person. You may assume the person's
area to be 1.50 m high by 0.50 m wide. Will this person be knocked off his feet if
he is not holding on to any support? State any further assumptions you have made
regarding the person and the ground.

Homework Equations

Density and momentum

The Attempt at a Solution

Here's how I derive the equation for the force. To me, it looked like nonsense, but the units checks out. I am not sure if what I am doing is correct and if it is, I am not sure how to justify it.

Density of air is

\rho=\frac{M}{V}

where M is mass and V is volume. We know that volume is Area x length. In this case, area is the area of the person and the length is speed of the wind v multiplied by the time the wind is hitting the person.

So far we have

\rho Al=M;

\rho Avt=M

We know that momentum is Mv, so we multiply v on both sides of the equation. We also know that Force is rate of change of momentum. So if we take the t as dt and divide both sides with that, we get

\rho Av^2=\frac{Mv}{dt}

And finally,

F=\rho Av^2

This gives me a numerical value of 1565N. This force will of course cause the person to be knocked of his feet.

Is it correct to say I am assuming that all the momentum of the wind is transferred to the person?

 

[Stonebridge]

Homework Statement

During a hurricane, winds can whip horizontally at speeds of 150 km/h. Given that
the density of air is 1.20 kg/m3 and the wind after striking the person is brought to
rest, calculate the force of the wind on the person. You may assume the person's
area to be 1.50 m high by 0.50 m wide. Will this person be knocked off his feet if
he is not holding on to any support? State any further assumptions you have made
regarding the person and the ground.

Homework Equations

Density and momentum

The Attempt at a Solution

Here's how I derive the equation for the force. To me, it looked like nonsense, but the units checks out. I am not sure if what I am doing is correct and if it is, I am not sure how to justify it.

Density of air is

\rho=\frac{M}{V}

where M is mass and V is volume. We know that volume is Area x length. In this case, area is the area of the person and the length is speed of the wind v multiplied by the time the wind is hitting the person.

So far we have

\rho Al=M;

\rho Avt=M

We know that momentum is Mv, so we multiply v on both sides of the equation. We also know that Force is rate of change of momentum. So if we take the t as dt and divide both sides with that, we get

\rho Av^2=\frac{Mv}{dt}

And finally,

F=\rho Av^2

This gives me a numerical value of 1565N. This force will of course cause the person to be knocked of his feet.

Is it correct to say I am assuming that all the momentum of the wind is transferred to the person?

That is the correct answer for the force assuming that the momentum of the wind is transferred to the person.
The key wording in the question being that the wind is "brought to rest".
That would be unlikely in reality.
1565N is the weight of a mass of about 160kg.
An adult male can have a mass of anywhere from about 70kg upwards.
What assumptions can you make to determine if he will be blown over or not?