Projectile motion w/drag equations

In summary: It's not that simple as integrating will result in an equation that is not equal. You would have to solve for each variable to get an equation that is equal.
  • #1
Mag|cK
36
0
Hello. How do we derive:
Maximum distance of a projectile as a function of initial velocity and launch angle. [taking into account the air resistance] Assuming mass of projectile, air viscosity, drag coefficient, etc are constants.

Is there any sites giving the steps?
 
Physics news on Phys.org
  • #2
I have another question: I have researched that
for horizontal component (x), Fdrag= -0.5 C p A V Vx
for vertical component (y), Fdrag= -0.5 C p A V Vy
where V=square root(Vx^2+Vy^2)

Isn't Fdrag for horizontal component is simply= -0.5 C p A Vx^2
and for vertical component, Fdrag = -0.5 C p A Vy^2 ?
 
  • #3
The horizontal and vertical drag components will depend on the angle, surely ?
Can you find the distance flown when there is no drag ?
 
  • #4
yes i can. I have found the answer to my second post. Any answers on my first post?
 
  • #5
Hm, I don't see any distances in your answers, or angles. Shouldn't you consider a projectile thrown with initial angle theta, and velocity V ?

This might help -
http://physics.gmu.edu/~amin/phys251/Topics/ScientificComputing/sprojectileMotion.html
 
Last edited by a moderator:
  • #6
I think you are talking about projectile motion without air resistance?
Nonetheless, in projectile with air resistance, is it possible to obtain an equation of maximum distance as a function of initial velocity and launch angle (as the variables only)?
 
  • #7
A good approach would be to write the equation for the maximum distance as a function of V and theta, then add terms to the derivation to account for air resistance. To get the maximum range you work out the time the projectile is in the air from the vertical component, then multiply that time by the horizontal velocity. If there's air resistance, the time in the air becomes a tricky calculation.

If there's air resistance it will enter the formula.
 
  • #8
It's not that simple as horizontal velocity is changing due to air drag.
 
  • #9
And btw why is this thread moved? I really don't think this is introductory physics.
 
  • #10
Who said it was simple ? The drag is obviously going to affect both components. Why don't you have a shot at it ? I'm willing to help if you show some work. You could also try looking for notes on the web.
 
  • #11
I want to integrate

dVx/dt = -(D v Vx) / m

but v=sqrt(Vx^2+Vy^2)
I have tried substitution method to eliminate Vy, but it's not possible (I have done this for hours).

My question is,
Is it even possible to come up with an equation of horizontal distance as a function of launch angle, initial velocity and time?
 
  • #12
Mag|cK said:
Is it even possible to come up with an equation of horizontal distance as a function of launch angle, initial velocity and time?
I cannot say for sure no, but I didn't see any such function so far.

However, I wonder why you need the closed form? If it is a goal unto itself, well... But if you need to compute this as part of some other problem, numeric solution can be easily constructed in many ways. The details are nicely presented in the link given by Mentz114.

--
Chusslove Illich (Часлав Илић)
 
  • #13
Well I have a set of equations, only based on that Fdrag= -0.5 c p a v^2. I just divide into horizontal and vertical component, mainly the force and acceleration. However to find the function of horizontal distance, I believe I have to integrate dVx/dt and then integrate it again.

While combining the equations (hoping that I can eliminate the unknown variable), I found a strange inconsistency in the equations:

(A=acceleration)

Fx= -D v^2 cosX = -D v Vx

Fy= -D v^2 sinX - mg = -D v Vy -mg

F=sqrt(Fx^2+Fy^2)
A=dv/dt= [sqrt(Fx^2+Fy^2)] /m ---------------[1]

v=sqrt(Vx^2+Vy^2)
dv/dt= (Vx Ax + Vy Ay)/ sqrt(Vx^2+Vy^2) --------[2]

Solving [1]=[2], they are not equal and after solving [1]=[2], I found that Vx=0 for [1] and [2] to be equal. Yet Vx is obviously not 0.

Can anyone explain why? I know they are time-consuming to solve (not difficult though). Thnx.
 
  • #14
Your expression [2] is not an equation, but an indentity: left hand side is acceleration component in direction of velocity, right hand side is the same, only using scalar product formulation.

Equation [1] is wrong: left hand side is same as in [2], but right hand side is full acceleration, not only the component in velocity direction. The corrected [1] would be alike [2]:
[tex]
\frac{dv}{dt} = \frac{v_x F_x + v_y F_y}{\sqrt{v_x^2 + v_y^2}} \frac{1}{m}
[/tex]

But with [2] being identity, [1] and [2] would only assert:
[tex]
a_x = \frac{F_x}{m}, \quad a_y = \frac{F_y}{m}
[/tex]
and you're back to square one.

--
Chusslove Illich (Часлав Илић)
 
  • #15
What's the difference between identity and equation?
So it's not possible to integrate and come up with distance equation?
 
  • #16
Mag|cK said:
What's the difference between identity and equation?
(I may not be using proper mathematical terms, so sorry for any confusion.)

Identity is something that is true by definition, so it cannot provide any links between the independent quantities in a particular problem. Equation is a constraint on the problem imposed from outside, e.g. by physical laws. Thus, your expression [1] comes from 2nd Newton law (an equation), whereas [2] comes from definitions of tangential acceleration and vector product (an identity).

Mag|cK said:
So it's not possible to integrate and come up with distance equation?
I cannot claim that it is not possible (math is not my strong point), I just haven't seen any such integration.

--
Chusslove Illich (Часлав Илић)
 

What is projectile motion with drag?

Projectile motion with drag is a type of motion in which an object is launched into the air and experiences the effects of air resistance or drag. This results in a curved path rather than a straight line.

What are the equations for projectile motion with drag?

The equations for projectile motion with drag include the equations for horizontal and vertical motion, as well as the equation for drag force. The horizontal equation is x = v0x * t, the vertical equation is y = v0y * t - 1/2 * g * t^2, and the equation for drag force is Fd = -1/2 * p * v^2 * Cd * A, where p is the density of the fluid, v is the velocity of the object, Cd is the drag coefficient, and A is the cross-sectional area of the object.

How does air resistance affect projectile motion?

Air resistance or drag affects projectile motion by slowing down the object and changing its trajectory. As the object moves through the air, it experiences a force in the opposite direction of its motion, causing it to decelerate and deviate from its original path.

What factors affect the trajectory of a projectile with drag?

The trajectory of a projectile with drag is affected by several factors, including the initial velocity of the object, the angle of launch, the mass and shape of the object, the density of the fluid, and the drag coefficient. These factors can alter the amount of drag force acting on the object, thus changing its trajectory.

How is projectile motion with drag used in real life?

Projectile motion with drag is used in real life in various fields such as sports, engineering, and physics. It is used to predict the trajectory of a thrown or launched object, design aerodynamic structures, and study the effects of air resistance on objects in motion.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
689
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
387
  • Introductory Physics Homework Help
Replies
7
Views
2K
Replies
2
Views
2K
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
20K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top