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.
$\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$

Question

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.

\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

Quantrex Academy · Accepted Answer

Correct answer: A. 4.2

Time for first drop to fall 5 m: 5 = (1)/(2) × 10 × t^2 gives t = 1 s. Since the first drop strikes when the sixth begins, there are 5 intervals: each interval T = (1)/(5) = 0.2 s. When drop 1 hits the ground (at t = 1 s), drop 4 has been falling for t_4 = 1 - 3 × 0.2 = 0.4 s. Distance fallen by drop 4: s_4 = (1)/(2) × 10 × (0.4)^2 = 0.8 m. Height of drop 4 above ground: 5 - 0.8 = 4.2 m.

Water drops fall from a tap on the floor, 5 m below, at regular...

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