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Grade 11Mechanics

Water drips from the nozzle of a shower onto the floor 2m below. The drops fall at regular intervals of time, the first drop striking the floor at the instant the fourth drop begins to fall. What is the location of the second drop from the floor?

Profile image of Aditya
7 Years agoGrade 11
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1 Answer

Profile image of Arun
ApprovedApproved Tutor Answer7 Years ago
Dear Aditya
 
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