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Q.

A particle is projected vertically upward with speed u = 10 m/s. During its journey air applies a force of -0.2v2 on the particle. What is the maximum height attained by the particle [m = 2kg]

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a

5 ln (2)

b

3ln (2)

c

2ln (2)

d

10 ln (2)

answer is A.

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Detailed Solution

a=−vdvdy=−g−0.2v22⇒dy=−10v dv100+v2⇒ymax=−5∫1002v dv100+v2⇒ymax=−5ln100+v2100= 5 ln [2]
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