A particle falls under constant
gravity in a resisting medium whose
resistance varies as the square of the
velocity. Then find the velocity of
the particle.
Answers
Answer:We are here probably considering a small sphere falling slowly through a viscous liquid, with laminar flow around the sphere, rather than a skydiver hurtling through the air. In the latter case, the airflow is likely to be highly turbulent and the resistance proportional to a higher power of the speed than the first.
We'll use the symbol y for the distance fallen. That is to say, we measure y downwards from the starting point. The equation of motion is
y¨=g−γv,(6.3.8)
where g is the gravitational acceleration.
The body reaches a constant speed when y¨ becomes zero. This occurs at a speed v^=gγ , which is called the terminal speed.
To obtain the first time integral, we write the equation of motion as
dvdt=γ(v^−v)(6.3.9)
or
dvv^−v=γdt.(6.3.10)
dvv−v^=−γdt.(6.3.11)
DON'T! In the middle of an exam, while covering this derivation that you know so well, you can suddenly find yourself in inextricable difficulties. The thing to note is this. If you look at the left hand side of the equation, you will anticipate that a logarithm will appear when you integrate it. Keep the denominator positive! Some mathematicians may know the meaning of the logarithm of a negative number, but most of us ordinary mortals do not - so keep the denominator positive!
With initial condition v=0 when t=0 , the first time integral becomes
Step-by-step explanation: