With what time interval did two drops come off the drain pipe if two seconds after

With what time interval did two drops come off the drain pipe if two seconds after the fall of the second drop the distance between the drops is 25 m?

Let’s point the x-axis down. The origin is at the point of the initial position of the drops.

Let, before the separation of the second drop, the first one has already fallen during the time interval t1. The coordinate of the first drop in t2 = 2 s after the separation of the second drop:

x1 = g (t1 + t2) ^ 2) / 2,

where g is the acceleration due to gravity.

Coordinate of the second drop 2 s after separation:

x2 = gt2 ^ 2/2;

x1 – x2 = 25 m;

g (t1 + t2) ^ 2) / 2 – gt2 ^ 2/2 = 25 m;

g (t1 ^ 2 + 2t1t2 + t2 ^ 2 – t2 ^ 2) = 50 m;

g (t1 ^ 2 + 2t1t2) = 50;

10 * t1 ^ 2 + 40 * t1 – 50 = 0.

Find the positive root of the equation:

t1 = (-40 + √ (1600 + 4 * 10 * 50)) / 20 = (-40 + 60) / 20 = 1 s.

Answer: the interval between droplet separation was 1 s.