Question

Do the algebraic solution first before plugging in the values.

3. Water drops from the nozzle of a shower onto the floor 81 inches below. The drops fall at regular intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. Find the location of the individual drops when a drop strikes the floor.

Answer 1

Let the first drop fall at t = 0

as the time interval is regular

let second drop falls at t = t

let third drop falls at t = 2t

let fourth drop falls at t = 3t

The height between nozzle and floor is h

therefore, from kinematics

h = 1/2at²

note that drop starts from rest.

t = sqrt (2h / g)

as the second drop had spent time 2t , we have

h' = 1/2g(2t)²

h' = 2gt²

notice that as the first drop hits the floor, the fourth drop begins to fall

so,

t² = 1/9 * 2h/g

so,

h' = 2g * 1/9 * 2h/g

h' = 4/9 * h

similarly, for third drop,

h'' = 1/2 * g * t²

h'' = 1/2 * g * 1/9 * 2h/g

h'' = h/9

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therefore, distance between nozzle and second drop is 4h/9

therefore, distance between nozzle and third drop is h/9

Now, just put the value of given h = 81 inches and solve,

Therefore,

distance between nozzle and second drop = 4*81 / 9 = 36 inches

distance between nozzle and third drop = 81 / 9 = 9 inches