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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a
T1, T2, T3
b
T2, T3, T1
c
T2, T1, T3
d
T1, T3, T2
answer is D.
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Detailed Solution
The first body will go up and after a time interval of , it will again corss the top of the tower with a velocity of u = 40m/s and then after a time T2, it will strike the ground. T1 = (8 + T2)sec. Also since the third body is released from rest, its time of flight T3 must be greater than T2, so option (4) is the correct option.