Projectile with air resistance

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SUMMARY

The discussion centers on calculating the greatest height reached by a particle projected vertically upwards with an initial speed U, while considering air resistance modeled as a retarding force of mkv². The participants confirm that the correct expression for the maximum height is (U²)/(2(g + kv²)). To find the speed W upon return to the original projection point, the participants emphasize the need for integration techniques due to the non-linear nature of the acceleration caused by air resistance. The final approach involves solving the differential equation dv/dt = -g - kv² and integrating to find both velocity and height.

PREREQUISITES
  • Understanding of kinematics and dynamics, particularly with non-linear forces.
  • Familiarity with differential equations and integration techniques.
  • Knowledge of the work-energy theorem and its application in mechanics.
  • Experience with calculus, specifically in solving motion problems involving variable acceleration.
NEXT STEPS
  • Study the integration of non-linear differential equations, focusing on the equation dv/dt = -g - kv².
  • Learn about the work-energy theorem and how it applies to systems with variable forces.
  • Explore numerical integration methods for solving differential equations using software like Excel or Maple.
  • Investigate the effects of air resistance on projectile motion in greater detail, including linear vs. quadratic drag forces.
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on mechanics and dynamics, as well as anyone interested in the effects of air resistance on projectile motion.

  • #31
Jess1986 said:
oh i see now. my differential equation skills are not too great. So i now integrate both sides?

Sorry, I had left a factor of "m" on the left side which should not have been there.

You can't integrate since v(t) is an unknown function. The fact that it is squared makes this a nonlinear equation which is tough. I am trying to guess the solution but I don't see it right away, sorry. :frown:
(maybe Maple could give you the answer).
 
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  • #32
thanks, you have still helped me get on the right track. once i find v do you know where i should go from there?
 
  • #33
Jess1986 said:
thanks, you have still helped me get on the right track. once i find v do you know where i should go from there?

In principle, here is what you have to do:

Sokve for v(t) with the initial condition v(0) = U.

set v(t)= 0 to find the time to get to the highest point.

Integrate v(t) to find y(t) (with the initial condition y(0)= 0). Plug in the time from the first part to find the max height reached.

Now start again, but now solving dv/dt = +g - k v^2 with the initial condition v(o) = 0.

Integrate v(t) to find y(t) with the initial condition being the max height.

Find the time at which y is zero again.

Plug back that time in the formula for v(t) on the way down.

It's that easy :biggrin:
 
  • #34
thanks very much for all your help. il give solving the equation in maple a go tomorrow and see how it goes!
 

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