How Do You Correctly Model Projectile Motion with Drag?

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Discussion Overview

The discussion revolves around modeling projectile motion with drag, specifically addressing the equations of motion for a projectile under the influence of a drag force. Participants are exploring the integration of initial conditions and the treatment of velocity as a variable in the context of this problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents an equation for net acceleration that includes drag and gravity, seeking to derive position functions over time.
  • Another participant suggests that horizontal and vertical velocities should be treated as variables rather than constants during integration.
  • There is a question about whether to plug in initial values or to keep initial velocity as a variable during the integration process.
  • A later reply emphasizes that the horizontal velocity equation should account for the drag force as a variable dependent on instantaneous velocity.

Areas of Agreement / Disagreement

Participants generally agree that velocities should be treated as variables during the integration process, but there is no consensus on the correct approach to integrating the equations or the treatment of initial conditions.

Contextual Notes

Participants express uncertainty about the integration process and the implications of treating velocities as variables, indicating potential limitations in their understanding of the mathematical steps involved.

Freyster98
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Homework Statement



If we include a crude model for the drag force in which the net acceleration on the ball kicked by a football player is given by:
a = (-k*vx)i + (-g-k*vy)j.
Derive the equations describing the horizontal and vertical positions as functions of time.
k=.031 (1/s), vo=69 (ft/s), [tex]\Theta[/tex]0=45.

Homework Equations


The Attempt at a Solution



I solved for vx and vy using the information given (vx=v0cos[tex]\Theta[/tex], vy=v0sin[tex]\Theta[/tex] ) plugged these values, along with k, into the acceleration equation. I took the integral of both the horizontal and vertical components independently to get velocity, then integrated that to get the position. The problem is, the horizontal velocity comes out to be -1.51*t, and horizontal position is -.755*t^2, which is obviously wrong because it would be moving backwards the instant it is kicked. What am I doing wrong here? Should I not be solving for vx and vy, and leaving those as variables as well?
 
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Freyster98 said:
… What am I doing wrong here? Should I not be solving for vx and vy, and leaving those as variables as well?

Hi Freyster98! :smile:

erm :redface: … yes …

vx and vy are definitely variables …

the original equation says that the acceleration is -g vertically, and -k (the drag coefficient) times the instantaneous velocity horizontally.
 
Im working on the same problem. Do I plug in the initial values and integrate. or do i leave the initial velocity as v0 and then integrate?
 
musichael said:
Im working on the same problem. Do I plug in the initial values and integrate. or do i leave the initial velocity as v0 and then integrate?

Hi musichael! :smile:

Leave v0 until the end

vx and vy are variables …

integrate, and you will get a constant …

at that stage you use v0 to find what the constant is. :wink:
 
The horizontal velocity is [tex]v_x = v_0 cos \theta_0 + a_x t[/tex], where [tex]a_x = -k v_x[/tex] is a variable and not a constant as you used.
 

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