Q) A particle is thrown upwards from the ground. It experiences constant air resistance which produces a retardation of 2 m/s^2. Find the ratio of time of ascent to time of descent.

A)

v = u + at

0 = u + -(g + 2)t

t = u / (g + 2)

0 - u^2 = 2as

- u^2 = 2 *-(g + 2)s

s = u^2 / 2(g + 2)

s = 1/2 aT^2

u^2 / 2(g+2) = (1/2) * (g-2)T^2

u / sqrt(g^2 - 4) = T

So t/T = sqrt (g-2) / sqrt (g+2) = sqrt (8/12) = sqrt (2/3)

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## Saturday, November 1, 2008

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