Guy Jumping out of Building (Air Resistance)

[DLPhysics]

Homework Statement

A 72 kg person escapes from a burning building by jumping from a window 30 m above a
catching net. The acceleration of gravity is 9.81 m/s2. Assuming that air resistance is simply a
constant 88 N force on the person during the fall, determine the person’s velocity just before hitting the net.

Homework Equations

Fg (force of gravity) - Air resistance = (mass) (acceleration)

Vf^2 = Vi^2 + 2ax

The Attempt at a Solution

I think the only reason that I am getting the question wrong is because of the fact that I think that inital velocity of the man jumping out is 0. I'm pretty sure that I plugged in everything on the first equation correctly, but I'm geting frustrated.

 

[Andrew Mason]

Your formula for speed is not correct.

$$\Delta v = at$$

I get 8.6 m/sec^2 as the acceleration. You just have to find the time of fall.

AM

 

[Cryphonus]

Yes i totally agree with Mr.Mason...

There is some forces that is acting on the guy,,, draw a diagram to see that clearly (there is a gravitational force on him,and a constant force that is actually preventing him to fall with gravitational acceleration(9,8))

This means you cannot take the value of the acceleration as 9,8 while he is falling to the ground, because your net acceleration is the one with resistance force included.(I think that's where your mistake is)

After you get your net value for acceleration you can easily calculate total time of flight and therefore calculate the final velocity.(note that this is like a projectile motion, so you will both have horizontal and vertical velocities)

 

[ehild]

Homework Equations

Fg (force of gravity) - Air resistance = (mass) (acceleration)

Vf^2 = Vi^2 + 2ax

The Attempt at a Solution

I think the only reason that I am getting the question wrong is because of the fact that I think that inital velocity of the man jumping out is 0. I'm pretty sure that I plugged in everything on the first equation correctly, but I'm geting frustrated.

Your formula is correct. As no initial velocity is mentioned, it should be zero. What result have you got?

ehild

 

[Andrew Mason]

yes. ehild is correct. I see what you are doing. The change in kinetic energy is mv^2/2 which has to be equal to the work done by the net force. W= F x d = mass x acceleration x distance. The distance is the height through which he falls.

$$v_f^2 = 2ah$$

You just have to use the correct acceleration.

AM