A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject to air drag (a resistive force caused bythe air). The drag force on t

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A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject to air drag (a resistive force caused bythe air). The drag force on t

Ayaana Buck 2020-12-29 Answered

A rubber ball of mass m is dropped from a cliff. As theball falls. it is subject to air drag (a resistive force caused bythe air). The drag force on the ball has magnitude $b v^{2}$ , where b is a canstant drag coefficient andv is the instantaneous speed of the ball. The dragcoefficient b is directly proportional to the cross-sectional areaof the ball and the density of the air and does not depend on themass of the ball. As the ball falls, its speedapproaches a constant value called the terminal speed.
a. Write, but do Not solve, a differentialequation for the instantaneous speed v of the ball in terms of timet, the given quantities quantities, and fundamentalconstants.
b. Determine the terminal speed vt interms of the given quantities and fundamental constants.
c. Detemine the energy dissipated by the dragforce during the fall if the ball is released at height h andreaches its reminal speed before hitting the ground, in terms ofthe given quantities and fundamental constants.

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unett

Answered 2020-12-30 Author has 119 answers

$σ F = m a$
$w - F_{η} = m a$
$m g - b v^{2} = m a$
$m g - k A ρ v^{2} = m a$ where k is the proportionalityconstant
$a = \frac{d v}{d t} = g - (\frac{k A ρ}{m}) v^{2}$
$\frac{d v}{d t} + \frac{k A ρ}{m} v^{2} = g$
I don't know if this is correct. The derivativeis with respect to time t, but is it considered "in terms oftime t?"
(b) When the terminal velocity is reached,there is no acceleration
$m g - k A ρ v^{2} = m a = 0$
$v = \sqrt{\frac{m g}{k A ρ}}$

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