How Does Air Resistance Affect Projectile Motion?

Your Name]In summary, the conversation revolves around finding the time and maximum height of a particle projected vertically with a given initial velocity under the influence of a drag force. Using equations for motion and integration, the solutions for a and b were found. The answers were then checked by taking the limit as the drag force approaches zero, which required the use of L'Hopital's rule. Finally, numerical evaluations were suggested to further understand the behavior of the particle with different initial velocities.
  • #1
pimpalicous
16
0

Homework Statement


A particle is projected vertically upward with a speed v0 under the influence of a drag force that is proportional to the particle's speed. As the particle falls back down, its terminal speed is vt.
(a) Find the time it takes to reach ymax

(b) Find ymax

(c)Show that your answers for a and b make sense by taking the limit where the drag force approaches zero. Do they reduce to the familiar eqns of 1st semester physics?

(d) evaluate a and b numerically for blah blah blah


Homework Equations


Fd=-k*m*v


The Attempt at a Solution



I got solutions for a and b pretty easily.

For a I wrote:
-mg-kmv=m[tex]\frac{dv}{dt}[/tex]
I separated and integrated

-[tex]\int dt[/tex]=[tex]\int dv/(g+kv)[/tex]
then
-t=[tex]\frac{1}{k}[/tex]*ln[[tex]\frac{g+kv}{g+kv_{0}}[/tex]]

I got a speed of
v=-[tex]\frac{-g}{k}[/tex]+([tex]\frac{g}{k}[/tex]+v0)e-kt

I set this equal to zero and solved for t to get

[tex]\Large t[/tex]=[tex]\frac{1}{k}[/tex]*ln[[tex]\frac{g+v_{0}k}{g}[/tex]]

For part b, I separated the v equation and integrated to find y

y(t)=-[tex]\frac{gt}{k}[/tex]+([tex]\frac{g}{k^{2}}[/tex] +[tex]\frac{v_{0}}{k}[/tex])[-e^(-kt) +1]

I plugged in the equation for t and got

ymax=-[tex]\frac{g}{k^2}[/tex]*ln[1+[tex]\frac{v_{0}k}{g}[/tex]]+([tex]\frac{g}{k^2}[/tex] + [tex]\frac{v_{0}}{k}[/tex])([tex]\frac{-g}{g+v_{0}k}[/tex] +1)

I checked them in c by taking the limit of t and ymax as k approached zero.

The t checked perfectly. I got t=[tex]\frac{v_{0}}{g}[/tex].

The ymax didn't go so well.

ymax=-[tex]\frac{g}{k^2}[/tex]*ln[1+[tex]\frac{v_{0}k}{g}[/tex]]+([tex]\frac{g}{k^2}[/tex] + [tex]\frac{v_{0}}{k}[/tex])(-g/(g+0) +1)

I used the approximation ln(1+x)=x-x^2/2 where v0k/g=x

ymax=-[tex]\frac{g}{k^2}[/tex][[tex]\frac{v_{0}k}{g}[/tex]-([tex]\frac{v_{0}^2*k^2}{2g^2}[/tex]]

it reduced to

ymax=-[tex]\frac{v_{0}}{k}[/tex]+[tex]\frac{v_{0}^2}{2g}[/tex]

The second term is what I'm looking for but the first term obviously goes to infinity and can't be right. Where did I make a mistake?
 
Physics news on Phys.org
  • #2


Thank you for your post. It seems like you have made some good progress in solving this problem. Here are some suggestions and clarifications on your solution:

a) To find the time it takes to reach ymax, you can set y=ymax in your equation for y(t) and then solve for t. This will give you the same result as you have for t in your solution.

b) To find ymax, you can use the equation for y(t) and set t=0 (since the particle starts at the ground) to find the initial height. Then, you can set v=0 (terminal velocity) and solve for y to find ymax. This will give you the same result as you have for ymax in your solution.

c) In order to take the limit as k approaches zero, you need to use L'Hopital's rule. This will help you evaluate the limit of the natural logarithm as k approaches zero. Also, note that your expression for ymax should not have a negative sign in front, since the particle is moving upwards initially.

d) To evaluate a and b numerically, you can plug in values for m, g, k, and v0. Then, you can use a calculator or computer program to solve for t and ymax. It might also be helpful to plot the graphs of t vs. v0 and ymax vs. v0 to see how they vary with different initial velocities.

Overall, your approach to solving this problem is correct, but there are some small mistakes and clarifications needed. Keep up the good work and feel free to ask for further assistance if needed.
 

FAQ: How Does Air Resistance Affect Projectile Motion?

What is the definition of 1-D air resistance problem?

The 1-D air resistance problem is a physics problem that involves calculating the effects of air resistance on an object moving through one dimension (such as horizontally or vertically).

How is air resistance calculated in a 1-D air resistance problem?

In a 1-D air resistance problem, air resistance is typically calculated using the formula F = 0.5 * C * ρ * A * v^2, where F is the air resistance force, C is the drag coefficient, ρ is the density of air, A is the cross-sectional area of the object, and v is the velocity of the object.

What factors affect the magnitude of air resistance in a 1-D air resistance problem?

The magnitude of air resistance in a 1-D air resistance problem is affected by several factors, including the shape and size of the object, the density of the air, and the velocity of the object.

How does air resistance impact the motion of an object in a 1-D air resistance problem?

In a 1-D air resistance problem, air resistance acts in the opposite direction of the object's motion, reducing its acceleration and velocity over time. This results in a slower and shorter trajectory compared to an object without air resistance.

Are there any real-world applications of 1-D air resistance problems?

Yes, 1-D air resistance problems have many real-world applications, such as in the design of airplanes, cars, and other vehicles, as well as in sports like cycling and skiing. Understanding air resistance is important in optimizing the efficiency and performance of these objects and activities.

Back
Top