Terminal velocity with an initial velocity?

[glueball8]

How could I find how long (time) it takes to reach terminal velocity with an initial velocity?

http://www.geocities.com/operation_rising_star/mis.tve.htm <- does this make sense?

using 1/2pV^2*C*A=ma

Thanks in advance.

 

[flatmaster]

As a quick answer, you only approach terminal velocity asymtotically, that is, you could ask when you reach 95% of terminal velocity, but technically when you reach terminal velocity

 

[glueball8]

As a quick answer, you only approach terminal velocity asymtotically, that is, you could ask when you reach 95% of terminal velocity, but technically when you reach terminal velocity

you mean you can never really reach terminal velocity, everyone keeps telling me that.

But I have a questions that ask when it reaches terminal velocity and how long does it take for the object to fall to the ground.

What's the long answer? lol

 

[flatmaster]

That page you cyted is no good. He mixed up free-fall acceleration and drag acceleration.

He got the drag force right though.


Fdrag = .5pCAv^2

Where,

v is velocity (m/s)

C is the object's Drag Coefficient.

A is the object's cross sectional area (),

p and is the density of air (assume 1.25 kg/m3)


Ok. What are the forces acting on the object?


F=ma

Gravity and the drag force. Assume up is the positive direction.

.5pCAv^2 - mg = ma


Here, you have a differential equation. Knowing a = dv/dt


B(v)^2 - D = G dv/dt

Is your differential equation to solve with B, D, and G representing new constants

 

[flatmaster]

In the real world, you can think of it as if you're falling at your terminal velocity after nearly reaching it. It's only mathematical rigor that says you can never reach your terminal velocity.

Calculating the time to reach the ground is easy for something with a large drag to mass ratio (skydiver with open parachute). You're quite simply moving at your terminal velocity. It's simply a constant velocity problem in the vertical (y) direction

where V is your terminal velocity and y is the total height to fall.

V = y/t

t = y * V


However, if you have something with a large terminal velocity, the total time will be longer than this because early in the fall, the object was moving slower than it's terminal velocity

 

[glueball8]

That page you cyted is no good. He mixed up free-fall acceleration and drag acceleration.

He got the drag force right though.


Fdrag = .5pCAv^2

Where,

v is velocity (m/s)

C is the object's Drag Coefficient.

A is the object's cross sectional area (),

p and is the density of air (assume 1.25 kg/m3)


Ok. What are the forces acting on the object?


F=ma

Gravity and the drag force. Assume up is the positive direction.

.5pCAv^2 - mg = ma


Here, you have a differential equation. Knowing a = dv/dt


B(v)^2 - D = G dv/dt

Is your differential equation to solve with B, D, and G representing new constants

Thank you for your reply.

The terminal velocity is 41.4m/s and its only a height of 200m

Hmmm I'm in high school so I'm not really sure differential equation. Did you just take the derivative?

"A simplified real world example of a differential equation is modeling the acceleration of a ball falling through the air (considering only gravity and air resistance). The ball's acceleration towards the ground is the acceleration due to gravity minus the acceleration due to air resistance. Gravity is a constant but air resistance is proportional to the ball's velocity. This means the ball's acceleration is dependent on its velocity. Because acceleration is the derivative of velocity, solving this problem requires a differential equation."

 

[Nick89]

A differential equation is an equation where the derivative of a variable depends on that same variable in one way or another.

For example:
\frac{dy}{dx} = 2y

What you can do in this case is called separation of variables: you bring all the y's to the left and all the x's to the right (or the other way around):
\frac{dy}{y} = 2\, dx

Now, you can integrate both sides:
\int \frac{dy}{y} = \int 2\, dx
\ln y = 2x + C

Then you can solve for y = y(x).