Linear motion question -- Drops falling from a dripping faucet

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Russ Morgan
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Hello, I am a new member looking for the answer to a question I recently had on an exam. I will not know if I got it right for up to 6 weeks so am curious.
Question is: A faucet drips water at 5 drops per second. calculate the distance in metres between the first and second drop after the first drop reaches 3 metres per second.I was a bit rushed so simply used v-u/a =t for time for first drop to reach 3 m/s
Then divided 5 drop per second to get 0.2. then subtracted the 0.2 from time in first equation, then used the new time in this formula s=(v+u/2)t.
I feel like this was too simple?

Thanks
Russ
 

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I suppose we are assuming no air friction... An we'll take G=10m/s2.
The first drop will reach 3m/s in time (3m/s)/G = 0.3 seconds.
So I would calculate how far that drop fell in that time and how far the second drop fell in 0.2 seconds greater than that time.
 
.Scott said:
So I would calculate how far that drop fell in that time and how far the second drop fell in 0.2 seconds greater than that time.
Did you mean less than that time? The second drop is in the air 0.2 s less than the first drop.
 
kuruman said:
Did you mean less than that time? The second drop is in the air 0.2 s less than the first drop.
I was trying to assist - not spell out the entire calculation.