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]

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

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]

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

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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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Detailed Solution

$a = - v \frac{dv}{dy} = - g - \frac{0 .2 v^{2}}{2}$
$\Rightarrow dy = - \frac{10 v dv}{100 + v^{2}}$
$\Rightarrow y_{\max} = - 5 \int_{10}^{0} \frac{2 v dv}{100 + v^{2}}$
$\Rightarrow y_{\max} = - 5 {[\ln [100 + v^{2}]]}_{10}^{0}$
= 5 ln [2]

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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 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]