Question

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?

Answer 1

height of the nozzle above floor = 181 cm = 1.81 m

the time taken by the first drop to fall through this distance can be calculated

by using the formula   S = u*t + (1/2)*a*t²

1.81 = 0 + 0.5* 9.8 * t ²

                                t² = 1.81 / 4.9

t² = 0.369 s²

t = 0.607 s

when the first drop reaches the floor , the fourth one is about to fall,

the second and third drops are in air.

According to this problem, the drops are falling on the floor at regular intervals of time

that is the time difference between 1st and 2nd = time difference between 2nd and 3rd = time difference between 3rd and 4th

let this interval be t₀ , then   t₀ = (1/3 )*t

                                               = (1/3) * 0.607 s

                                               = 0.202 s

when the first drop has struck the floor , the 2nd drop has fallen for a time 0.404 s , the 3rd one has fallen for a time 0.202 s, while the forth one is just to fall.

(A)

Heigth through which 2nd drop fallen S₂ = (1/2 )*g*t₁²

                          here, t₁  = 2t₀  , then        S₂ = 0.5 * 9.8 * (0.404 s)²

                                                               S₂ = 0.7997 m

S₂ = 79.97 cm   

(B)

Heigth through which 3 rd drop fallen S₃ = (1/ 2)*g *t₂²         , here t ₂ = t _o

S₃ = 0.5 * 9.8 * (0.202 s)²

S₃ = 0.1999 m

S₃ = 19.99 cm