A stone is projected vertically up from the b | vedclass

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(C) When the stone moves up,both gravity $(mg)$ and the water drag force $(F_{drag})$ act downwards. Therefore,the net retarding force is $F_{up} = mg

Vedclass · Accepted Answer

$t_{up} < t_{down}$

Vedclass · Answer

(C) When the stone moves up,both gravity $(mg)$ and the water drag force $(F_{drag})$ act downwards. Therefore,the net retarding force is $F_{up} = mg + F_{drag}$,and the acceleration is $a_{up} = g + (F_{drag}/m)$.When the stone moves down,gravity acts downwards while the water drag force acts upwards. Therefore,the net accelerating force is $F_{down} = mg - F_{drag}$,and the acceleration is $a_{down} = g - (F_{drag}/m)$.Since $a_{up} > a_{down}$,the stone experiences a higher deceleration while moving up and a lower acceleration while moving down.For the same displacement $h$,using $h = \frac{1}{2}at^2$,we have $t = \sqrt{\frac{2h}{a}}$.Since $a_{up} > a_{down}$,it follows that $t_{up} < t_{down}$.

$A$ stone is projected vertically up from the bottom of a water tank. Assuming no water resistance,it will go up and come down in the same time. However,if water drag (resistance) is present,then the time it takes to go up,$t_{up}$,and the time it takes to come down,$t_{down}$,are related as:

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