Water drips from the nozzle of a shower onto the floor 181 cm below. The drops fall at regular (equal) intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. When the first drop strikes the floor, how far below the nozzle are the (a) second and (b) third drops?

If t be the time of fall of drops from a height of 181 s,

181 = 1/2 g t²

t = 6 s approx

There are 4 drops in vertical line, one touching the floor, one at 181 m height on the verge of falling down, and two drops in mid air.

The time interval between dripping of two consecutive drops

= 6 / 3

2 s

time duration of fall of second drop = 2 x 2 = 4 s

Position of drop below nozzle

= 1/2 g x 4²

= 78.4 m

b)

Time duration of fall of third drop

= 2 s

Position of drop below nozzle

= 1/2 9.8 x 2²

= 19.6 m