Vikas TU
Last Activity: 8 Years ago
Let the drops fall at time interval of t
Let the first, 2nd, 3rd and 4th drop fall at time 0, t, 2t, and 3t respectively.
Let height to which every drop falls = h
Time taken to fall = sqrt(2h/g)
Time interval between first drop and fourth drop = 3t
When first drop hits the floor, then 4th drop begins to fall.
Therefore, 3t = sqrt(2h/g)
Or, t = 1/3 * sqrt(2h/g) -----------(2)
2nd drop falls at t. Therefore, at time 3t, the second drop has spent time 3t - t = 2t.
Distance of 2nd drop from the nozzle = 1/2 * g * (2t)^2 = 1/2 * g * 4t^2
= 2gt^2
= 2g * 1/9 * 2h/g [using the value of t from (1)]
= 4/9 * h
= 4/9 * 2 m = 0.8889 meter.
3rd drop falls at 2t. Therefore, at time 3t, the second drop has spent time 3t - 2t = t.
2nd drop is 0.8889 m below the nozzle
