Water drips from the nozzle of a shower onto the floor 184 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?

Distance from the second drop is 81.5 cm

Distance from the third drop is 20.5 cm

Explanation:

Given total height = 184 cm = 1.84 m

By using second equation of motion we can find the total time of drop to reach the ground

s = vi + (1/2) a*t²

s stands for distance or height

vi stands for initial velocity

t stands for time

a stands for gravitational constant

s = 1.84 m

vi = 0 m/s

g = 9.8 m/s²

s = vi + (1/2) a*t²

1.84 = 0 + (1/2) * 9.8*t²

1.84 = 4.9 * t²

t = 0.612 s

The time interval for each drop is equal so 4 drops there are 3 time intervals. We can find the time interval by dividing the total time by 3.

time interval = total time / 3

= 0.612/3

= 0.204 s

Part a : Distance from the second drop

t=2 * time interval

t = 2 * 0.204

t = 0.408 s

s = vi + (1/2) a*t²

s = 0 + (1/2) 9.8 * (0.408) ²

s = 4.9*0.166

s =.815 m = 81.5 cm

Distance from the second drop is 81.5 cm

Part b : Distance from the third drop

t = time interval

t = 0.204 s

s = vi + (1/2) a*t²

s = 0 + 0.5*9.8 * (0.204) ²

s = 4.9*0.042

s = 0.205 m = 20.5 cm

Distance from the third drop is 20.5 cm