Vertical Motion with Quadratic Air Resistance

In summary, the baseball's maximum height is 7cm when it is in a vacuum and it is 15kg when it is in a real world.
  • #1
Rumble
2
0
Before I write anything, I want to apologize because I have no idea how to write equations on this website. This is my first post >.< Also, thank you for helping in advance!

Homework Statement


A baseball is thrown vertically up with speed vo and is subject to a quadratic drag with magnitude f(v) = cv2. Write down the equation of motion for the upward journey (measuring y vertically UP) and show that it can be rewritten as v(dot) = -g[1+(v/vter)2]. Use the "vdv/dx rule" to write v(dot) as vdv/dy and then solve the equation of motion by separating variables (put all terms involving v on one side and all terms involving y on the other). integrate both sides to give y in terms of v, and hence v as a function of y. Show that the baseball's maximum height is

ymax = [(vter)2/2g]*ln[ [ (vter)2 + (vo)2 ] / [(vter)2] ]

whew. If vo = 20m/s and the baseball has the parameters: mass m=.15kg and diameter D = 7cm, what is ymax? Compare with the value in a vacuum.


Homework Equations


Ok... Well first, in case you didn't get it, the vdv/dx rule is just that:

v(dot) = vdv/dx = (1/2)d(v2)/dx.

(only in this problem we just use y instead of x.)

Another formula that's important is the terminal velocity, which is
vter = sqrt(mg/c)


The Attempt at a Solution



Well, the first thing it asks is to write down the equation of motion. I'm a little unsure, but I think that it is :

m*v(dot) = -mg - cv2

which can be rearranged:
v(dot) = -g - cv2/m

and substituting c/m = g/(vter)2 in...
v(dot) = -g (1 + (v/vter)2)

so then we use the vdv/dx rule...
vdv = -g*dy*(1 + (v/vter)2)

and separating variables like it said,
vdv/(1 + (v/vter)2) = -gdy

But now I'm not sure what I'm supposed to do. When it said to separate variables, it said that I should put the terms with a y on one side and the terms with a v on the other, but... are there any terms with a y? Other than the dy? I also have no idea how to integrate this equation... Can anybody help me figure out the next few steps? Thank you again.

PS: is there a way to actually have it write v(dot) normally - as in, with a dot above the v?
 
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  • #2
Ok so I went ahead and tried to integrate vdv/(1 + (v/vter)2) = -gdy

On the left, I went from vo to v and on the right i went from 0 to y. This gave me (and watch out i switch the left and right sides here):

-gy = sqrt[ (1/vter2)*v2 + 1 ] / (1/vter2) from vo to v.

Now, vo = 0 at ymax, so you can plug those in, and you get

-gymax = (vter)2*sqrt[ (1/vter2)*v2 + 1 ] from vo to 0

which simplifies to

-gymax = vter2 * (1 - sqrt[ (vo/vter)2 +1 ]

and I have no idea how to make that into the given equation (with ln and stuff) that is shown in my first post.
 
  • #3
Rumble said:
vdv/(1 + (v/vter)2) = -gdy
-gy = sqrt[ (1/vter2)*v2 + 1 ] / (1/vter2)
No, that integration step with v is wrong. Please write it out in more detail.
 

FAQ: Vertical Motion with Quadratic Air Resistance

1. What is vertical motion with quadratic air resistance?

Vertical motion with quadratic air resistance is a type of motion where an object is moving vertically (up or down) while also experiencing air resistance. This type of motion is described by a quadratic equation that takes into account the object's mass, acceleration due to gravity, and the air resistance force.

2. How does air resistance affect vertical motion?

Air resistance is a force that opposes the motion of an object through air. In vertical motion, air resistance can affect the object's velocity and acceleration. As the object moves faster, the air resistance force also increases, causing the object's acceleration to decrease. This results in a more gradual decrease in velocity compared to when there is no air resistance.

3. How is the quadratic equation used to model vertical motion with air resistance?

The quadratic equation used to model this type of motion is:
y = y0 + v0t - (g/2m)t2 + (k/m)t2
where y is the object's vertical position, y0 is the initial position, v0 is the initial velocity, g is the acceleration due to gravity, k is the air resistance coefficient, and m is the object's mass. This equation takes into account both the effects of gravity and air resistance on the object's motion.

4. How does the mass of an object affect its vertical motion with air resistance?

The mass of an object affects its vertical motion with air resistance in two ways:
1) A heavier object will experience a greater gravitational force, causing it to accelerate faster towards the ground.
2) A heavier object will also experience a greater air resistance force, slowing it down more compared to a lighter object. This means that the mass of an object affects both its acceleration and terminal velocity in vertical motion with air resistance.

5. Are there any real-life applications of vertical motion with quadratic air resistance?

Yes, there are several real-life applications of this type of motion, such as skydiving, bungee jumping, and parachuting. In these activities, the motion of the person is affected by both gravity and air resistance. Understanding this type of motion is also important in fields such as engineering and physics for designing and analyzing structures and vehicles that move through air.

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