From the top of a tower two bodies are projected with the same initial speed of 40 ms^–1, first body vertically upwards and second body vertically downwards. A third body is freely released from the top of the tower. If their respective times of flights are T₁, T₂ and T₃ identify the correct descending order of the times of flights

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T₁, T₂, T₃

T₂, T₃, T₁

T₂, T₁, T₃

T₁, T₃, T₂

answer is D.

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Detailed Solution

according to the question we get,

the body which is projected upwards will be taking more time compare to the body which is projected downwards because the time is spent in covering the upward trajectory.

also, the body which released has zero initial velocity it will take more time than the body which is thrown downwards, and will take less time than the body which is thrown upwards.

therefore , T₁ > T₃ > T₂

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