From the top of a tower two bodies are projected with the same initial speed of $40 m s^{- 1}$ . First body is projected vertically upwards and second body is projected vertically downwards. A third body is freely released from the top of the tower. If their respective times of flight are $T_{1}$ , $T_{2}$ and $T_{3}$ , identify the correct descending order of the times of flight.

T_{1} > T_{2} > T_{3}

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T_{2} > T_{3} > T_{1}

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T_{2} > T_{1} > T_{3}

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T_{1} > T_{3} > T_{2}

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Solution

The correct option is C $T_{1} > T_{3} > T_{2}$
Clearly, for the body which is projected upwards will take more time than the body which is projected downwards as the time is spent in covering the upward trajectory.
And for the body which is just released has zero initial velocity. So it will take more time than the body which is thrown downwards and will take less time than the body which is thrown upwards.
Thus, $T_{1} > T_{3} > T_{2}$

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