JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Kinematics

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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 / s^{2}

)

(A)2.5
(B)4.0
(C)4.2✓ Correct
(D)3.8

Answer: (C)

Solution

Time for first drop to reach ground:

t = \frac{2 h}{g} = \frac{2 \times 5}{10} = 1

s

Since 6 drops are released in 1 second (first drop falls while sixth begins), the time interval between consecutive drops is

Δ t = \frac{1}{5} = 0.2

s.

When the first drop strikes the ground, the fourth drop has been falling for:

t_{4} = 1 - 3 (0.2) = 1 - 0.6 = 0.4

s

Distance fallen by fourth drop:

s = \frac{1}{2} g t_{4}^{2} = \frac{1}{2} \times 10 \times (0.4)^{2} = 0.8

m

Height from ground:

h = 5 - 0.8 = 4.2

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Kinematics

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