A particle is thrown upwards from ground. It experiences a constant resistance force which can produce a retardation of $2 m s^{- 2} .$ The ratio of time of ascent to time of descent is (g= $10 m / s^{2}$ )

1 : 1

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\sqrt{\frac{2}{3}}

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\frac{2}{3}

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\sqrt{\frac{3}{2}}

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Solution

The correct option is B $\sqrt{\frac{2}{3}}$
$v = u + a t$
$0 = u + - (g + 2) t$
$t = \frac{u}{g + 2}$
$0 - u^{2} = 2 a s$
$- u^{2} = 2 \times - (g + 2) s$
$s = u^{2} / 2 (g + 2)$
$s = \frac{1}{2} a T^{2}$
$\frac{u^{2}}{2 (g + 2)} = \frac{1}{2} \times (g - 2) T^{2}$
$\frac{u}{\sqrt{(} g^{2} - 4)} = T$
$\frac{t}{T} = \frac{\sqrt{g - 2}}{\sqrt{g + 2}} = \sqrt{\frac{8}{12}} = \sqrt{\frac{2}{3}}$

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