If the effects of atmospheric resistance are accounted for a fallinf body has an acceleration defined by the equation a=9.81[1-v^2(10^-4)] where v is in m\s and positive direction is downward

If the body is released downward from rest at a very high altitude determinethevelocity when t=3

Determine the bodys terminal or maximum attainable velocity as ( t goes to infinity)

Respuesta :

Answer:

A. Velocity = 28.61m/s

B. Maximum velocity = 100m/s

Explanation:

Remember that acceleration due to gravity, a is the change in velocity divided by the change in time

a(t) = dv/dt

This equation can be written as

dt = (1/a(t))dv

Integrating both sides

Integral(dt) = integral((1/a(t))dv)

Inputting a = 9.81[1 - v^2(10^4)]

Integral(dt) = integral((1/9.81[1 - v^2(10^-4)]dv)

t = 50/9.81 * Ln[(1 + 0.01v)/(1 - 0.01v)]

Making v the subject of formula,

v = (100 * [(exp(0.1962*t) - 1)/(exp(0.1962*t) + 1)])

A. t = 3s

v = (100 * [(exp(0.5886) - 1)/(exp(0.5886) + 1])

= 28.61 m/s

B. At t = infinity

Vmax = 100m/s

Attached the steps in the integration for better understanding.

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