An object of mass 5 , kg is thrown vertically | vedclass

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(B) During ascent,both gravity and air resistance act downwards. The net retarding force is $F_{up} = mg + f = (5 	imes 10) + 10 = 60 \, N$. The acce

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$\sqrt{2}: \sqrt{3}$

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(B) During ascent,both gravity and air resistance act downwards. The net retarding force is $F_{up} = mg + f = (5 	imes 10) + 10 = 60 \, N$. The acceleration during ascent is $a_{up} = F_{up} / m = 60 / 5 = 12 \, ms^{-2}$.During descent,gravity acts downwards and air resistance acts upwards. The net force is $F_{down} = mg - f = (5 	imes 10) - 10 = 40 \, N$. The acceleration during descent is $a_{down} = F_{down} / m = 40 / 5 = 8 \, ms^{-2}$.Let $h$ be the maximum height reached. For ascent,$h = \frac{1}{2} a_{up} t_1^2$,so $h = \frac{1}{2} 	imes 12 	imes t_1^2 = 6 t_1^2$.For descent,$h = \frac{1}{2} a_{down} t_2^2$,so $h = \frac{1}{2} 	imes 8 	imes t_2^2 = 4 t_2^2$.Equating the two expressions for $h$: $6 t_1^2 = 4 t_2^2$,which gives $\frac{t_1^2}{t_2^2} = \frac{4}{6} = \frac{2}{3}$.Therefore,the ratio of time of ascent to the time of descent is $\frac{t_1}{t_2} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$.

An object of mass $5 \, kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 \, N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to [Use $g = 10 \, ms^{-2}$]

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