A ball is thrown vertically upwards from the ground. If $T_{1}$ and $T_{2}$ are the respective time taken in going. up and coming down, and the air resistance is not ignored, then

T_{1} > T_{2}

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T_{1} = T_{2}

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T_{1} < T_{2}

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Solution

The correct option is C $T_{1} < T_{2}$
Since air resistance is not ignored, thus while going up, resistance acts opposite to velocity or in downwards direction. Thus net acceleration is in downwards direction and is equal to $a + g$ , which opposes motion. Thus time taken will reduce to bring it to stop. While coming down, r acts opposite to velocity , i.e. upwards, thus net force or acceleration is downwards $= a - g$ . Since acceleration is lower in this case, time will not decrease that much as compared to $a + g$ . Thus $T_{1}$ will be less than $T_{2}$

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A ball is thrown vertically upwards from the ground. If T1 and T2 are the respective time taken in going. up and coming down, and the air resistance is not ignored, then

A ball is thrown vertically upwards from the ground. If $T_{1}$ and $T_{2}$ are the respective time taken in going. up and coming down, and the air resistance is not ignored, then