Question

The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at the instant the first touches the ground. How far above the ground is the second drop at that instant?​

Answer 1

Explanation:

Let's find out ,

see the attachment for figure

Now as the drops are falling at regular intervals

let time between two drops be t .

If we add the total time (as per figure)

Total time for one drop = t+t = 2t

total distance = 5 m ..(given)

u (initial velocity) = 0 m/s

therefore,

s = ut + 1/2 at²

5 = 1/2 × 10 × (2t)²

5 = 5 × 4t²

t² = 1/4

t = 1/2

Thus each drop is released after 1/2 seconds each

now let's find x

x = ut + 1/2 gt²

x = 1/2 × 10 × (1/2)²

x = 5/4

x = 1.25 m

Therefore h = 5-x = 5 - 1.25 = 3.75m

Therefore the 2nd drop was at a height of 3.75 m from the ground !!!

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