water drop fall from a tap on the floor 5 M below at regular intervals of time the first drop is dragging The Frozen V begin to fall the height at which the third eye drop will be from ground at the instant win the first drop strike the ground will be

Height of tap = 5m and (g) = 10 m/sec².

For the first drop, 5 = ut +  gt ²

= (0 × t) +  × 10 t^{2 v}= 5t² or t² = 1 or t = 1.

It means 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 = 3.75 m.

Regards

Arun (askIITians forum expert)