JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Kinematics

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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.

(A)7.35 m✓ Correct
(B)2.45 m
(C)2.94 m
(D)4.18 m

Answer: (A)

Solution

The first drop reaches the floor in

t = \frac{2 h}{g} = \frac{2 \times 9.8}{9.8} = 2

s. The third drop leaves exactly as the first lands, so the drops are released at equal intervals of

\frac{t}{2} = \frac{2}{2}

s, and at that instant the second drop has been falling for one interval. It has fallen

s = \frac{1}{2} g (\frac{t}{2})^{2} = \frac{1}{2} \times 9.8 \times \frac{2}{4} = 2.45 m,

so the second drop is

9.8 - 2.45 = 7.35

m above the floor.

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Kinematics

Solution

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