5. Water drops from a tap at a regular interval of 2 s. When 1st drop is at floor, 4th drop is about to
leave. Find the position of 2nd drop from the floor at this instant.
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Height of tap = 5m and (g) = 10m/sec
2
For the first drop, 5=ut+gt
2
=(0×t)+×10t
2v
=5t
2
or t
2
=1 or t=1.
It mean that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.
Distance covered by the second drop in 0.5 sec.
=(0.5)
2
=1.25m
Therefore, distance of the second drop above the ground
=5−1.25=3.75m
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