Fluid Friction Question: Calculating Projectile Velocities with Air Resistance

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SUMMARY

This discussion focuses on calculating the velocities of three projectiles subject to air resistance at a distance of 5 meters. The projectiles have initial velocities of 121.632 m/s, 136.8 m/s, and 182.442 m/s, with a mass of 2.0x10^-4 kg, a cross-sectional area of 2.81x10^-5 m^2, and a drag coefficient of 0.47. The drag force is calculated using the equation Fa = 0.5 * D * v^2 * Cd * A, where D is the air density of 1.204 kg/m^3. The discussion emphasizes the need to derive a differential equation for velocity over time to accurately determine the projectile's velocity after traveling 5 meters.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with the drag equation (Fa = 0.5 * D * v^2 * Cd * A)
  • Knowledge of differential equations and their applications in physics
  • Basic concepts of projectile motion and air resistance
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  • Learn how to derive and solve differential equations for motion under drag forces
  • Study the application of numerical methods for solving projectile motion problems
  • Explore the impact of varying drag coefficients on projectile trajectories
  • Investigate computational tools for simulating projectile motion with air resistance
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Students in physics or engineering, educators teaching projectile motion, and anyone interested in the effects of air resistance on projectiles.

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Homework Statement



I'm trying to compare the velocities at 5 meters of three projectiles subject to air resistance: the first with an initial velocity of 121.632 m/s, the second with an initial velocity of 136.8m/s, and the third at 182.442 m/s.

All three projectiles have a mass of 2.0x10^-4 kg (m), a cross sectional area of 2.81x10^-5 m^2 (A), and drag coefficient of .47 (Cd). Density of air is assumed to be 1.204 kg/m^3.

Homework Equations



The equation I have been primarily been using is the drag equation Fa = 0.5Dv^2CdA

My initial plan was to use F = ma and Vf^2 = Vi^2 + 2a[itex]\Delta[/itex]x. However, I realized after doing these calculations that Fa changes with respect to velocity.

The Attempt at a Solution


This is my attempt to the solution of finding the velocity after 5 meters for 121.632 m/s.

Fa = -(0.5)(1.204)(121.632 m/s)^2(.47)(2.81E-5) = -1.2E-1

Which I then realized that only applied initially at launch.

So I tried taking the derivative with respect to time.

dFa = (.47)(2.81E-5)(dv/dt)

I am fairly stuck at the moment. Where do I go from now? If there is not enough information, what information do I need and what hints would you give to experimentally gain this information?
 
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Here is my suggestion.

Using the drag equation, Newton's second law and the fact that [itex]a = \frac{d v}{d t}[/itex] (where [itex]a[/itex] is the acceleration) you can find a differential equation for [itex]v[/itex]. Solving that, you can find [itex]v[/itex] at any time [itex]t[/itex]. Then, using the fact that [itex]v = \frac{d x}{d t}[/itex], you can find another differential equation for [itex]x[/itex]. Solving this, you can find the time at [itex]x = 5m[/itex]. Then you can plug in that time to the equation for [itex]v[/itex].

Let me know if it works.
 

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