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Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8 m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

1. 4.18 m
2. 2.94 m
3. 2.45 m
4. 7.35 m

Correct answer: 7.35 m

Solution

Time for the first drop to fall 9.8 m: t=sqrt(2h/g)=sqrt(2) s. Drops fall at equal interval T=t/2 (3rd starts as 1st lands). The 2nd drop has fallen for t/2 s: distance=0.5*9.8*(t/2)^2=2.45 m, so its height above the floor is 9.8-2.45=7.35 m.

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