Question

A particle of mass m is falling under the influence of gravity through a medium whose resistance equals μ times the velocity. If the particle be released from rest, show that distance fallen through time t is

Image ignouacademy-com--p-through-29659\frac{gm^2}{\mu ^2}\left \{ e^\frac{-\mu t}{m}-1+\frac{\mu t}{m} \right \}.

15 Feb 2018
Answer :
Word Count : 507
To solve this problem, we need to apply Newton's second law of motion to the particle falling through the medium with resistance proportional to velocity. ### Step 1: Set up the equation of motion For a particle of mass \( m \) falling under gravity with resistance proportional to velocity, the forces acting on the particle are: - The gravitational force: \( F_{\text{gravity}} = mg \) - The resistive force: \( F_{\text{resistance}} = -\mu v \), where \( \mu \) is a constant and \( v \) is the velocity of the particle. By Newton's second law, the sum of forces is equal to mass times acceleration: \[ ma = mg - \mu v \] Since acceleration \( a = \frac{dv}{dt} \), we can rewrite the equation as: \[ m \frac{dv}{dt} = mg - \mu v \] Dividing both sides by \( m \): \[ \frac{dv}{dt} =
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