Deriving the speed of a falling ball

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SUMMARY

The discussion focuses on deriving the speed of a falling baseball under the influence of gravity and air resistance, modeled by the drag force proportional to the square of the speed (f=Dv²). The terminal speed is established as vt = √(mg/D), where m is mass, g is gravitational acceleration, and D is the drag coefficient. The participants successfully apply Newton's second law to derive the equation for the speed at any time, ultimately isolating v as v = √(mg/D) * tanh(√(mg/D) * t).

PREREQUISITES
  • Understanding of Newton's second law of motion
  • Familiarity with drag force concepts in physics
  • Knowledge of hyperbolic functions, specifically tanh
  • Basic calculus for integration and differential equations
NEXT STEPS
  • Study the derivation of terminal velocity in fluid dynamics
  • Explore the applications of hyperbolic functions in physics
  • Learn about differential equations related to motion under resistance
  • Investigate the effects of varying drag coefficients on falling objects
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of falling objects and the effects of air resistance on motion.

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Homework Statement


You drop a baseball from the roof of a tall building. As the ball falls, the air exerts a drag force proportional to the square of the ball's speed (f=Dv2).
  1. (a) In a diagram, show the direction of motion and indicate, with the aid of vectors, all the forces acting on the ball.
  2. (b) Apply Newton's second laws second law and infer from the resulting equation the general properties of the motion.
  3. (c) Show that the ball acquires a terminal speed that is described by vt= \sqrt{\frac{mg}{D}}
  4. (d) Derive the equation for the speed at any time.

Homework Equations


\int((a2-x2)-1)dx = \frac{1}{a}arctanh(\frac{x}{a}), where tanh(x)=(ex-e-x)/(ex+e-x)= (e2x-1)/(e2x+1)

The Attempt at a Solution


In the attachment.

I'm at a block with the last step of part d. How do I isolate v from that? Any help is appreciated.
 

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tanh^{-1}(\frac{v\sqrt{D}}{\sqrt{mg}}) = \sqrt{\frac{mg}{D}}*t
...
Finally
v = \sqrt{\frac{mg}{D}}*tanh(\sqrt{\frac{mg}{D}}*t)
 
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Likes   Reactions: hhhh
Oh wow. That was pretty simple in retrospect... thanks.
 

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