Water drops fall from a tap on the floor, 5 m below, at regular intervals of time, the first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground, at the instant when the first drop strikes the ground is ____ m.
(g=10 m/s²)

Solution

Physics sir · 3 steps

Find the gap between drops from the first drop's 1 s fall, then see how long drop four has been falling.

1. Total fall time. A drop starts from rest and falls 5 m, so 5 = 1/2(10)t². That gives t = 1 s for the first drop.
2. Gap between drops. When drop 1 lands, drop 6 is just leaving the tap, so there are 5 equal gaps in that 1 s. Each gap is 1/5 = 0.2 s.

   Drop	Time spent falling	Distance fallen, 5t²	Height above floor
   1	1.0 s	5 m	0 (just landing)
   2	0.8 s	3.2 m	1.8 m
   3	0.6 s	1.8 m	3.2 m
   4	0.4 s	0.8 m	4.2 m
   5	0.2 s	0.2 m	4.8 m
   6	0	0	5 m (just leaving)

3. So. Drop 4 has been falling for 2 × 0.2 = 0.4 s and has dropped 5(0.4)² = 0.8 m. It is 5 - 0.8 = 4.2 m above the ground, which is option (c).

The idea to keep. For drops released at equal intervals, n drops in the air means n minus 1 gaps. Find each drop's time of fall and use h = 1/2gt², then subtract from the total height.

Answer from the NTA final answer key. Solution by PrepWiser.