A particle thrown up vertically reaches its highest point in time t1 and returns to the ground in time t2.The air resistance exerts a constant force on the particle opposite to its direction of motion, then

See figure For upward motion0=u−a1t1 or u=a1t1and h=ut1−12a1t12∴h=a1t1t1−12a1t12=12a1t12For downward motionh=12a2t2∴t2t1=a1a2=[g+(R/m)][g−(R/m)]or t1<t2

A particle thrown up vertically reaches its highest point in time t1 and returns to the ground in time t2.The air resistance exerts a constant force on the particle opposite to its direction of motion, then

See figure For upward motion0=u−a1t1 or u=a1t1and h=ut1−12a1t12∴h=a1t1t1−12a1t12=12a1t12For downward motionh=12a2t2∴t2t1=a1a2=[g+(R/m)][g−(R/m)]or t1<t2

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A particle thrown up vertically reaches its highest point in time t₁ and returns to the ground in time t₂.
The air resistance exerts a constant force on the particle opposite to its direction of motion, then

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$t_{1} > t_{2}$

$t_{1} = t_{2}$

$t_{1} < t_{2}$

none of above

answer is C.

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Detailed Solution

See figure

For upward motion
$0 = u - a_{1} t_{1} or u = a_{1} t_{1}$
$\begin{array}{ll} and h = {ut}_{1} - \frac{1}{2} a_{1} {t_{1}}^{2} \\ ∴ h = (a_{1} t_{1}) t_{1} - \frac{1}{2} a_{1} {t_{1}}^{2} = \frac{1}{2} a_{1} {t_{1}}^{2} \end{array}$
For downward motion
$\begin{array}{l} h = \frac{1}{2} a_{2} t_{2} \\ ∴ \frac{t_{2}}{t_{1}} = \sqrt{(\frac{a_{1}}{a_{2}})} = \sqrt{\frac{[g + (R / m)]}{[g - (R / m)]}} \\ or t_{1} < t_{2} \end{array}$