First slide

Rectilinear Motion

Question

Water drops fall from a tap on the floor 5 m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be from ground (at the instant when the first drop strikes the ground) will be $(g = 10 {ms}^{- 2})$

Moderate

A

1.25m
B

2.15m
C

2.75m
D

3.75m

Solution

By the time fifth water drop starts falling, the first water drop reaches the ground
$u = 0, h = \frac{1}{2} g t^{2} \Rightarrow 5 = \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 1 s$
Hence, the interval of falling of each water drop is
$\frac{1 s}{4} = 0.25 s$
When the fifth drop starts its journey towards ground, the third drop travels in air for
$0.25 + 0.25 = 0.5$
Therefore, height (distance) covered by third drop in air is
$h_{1} = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times (0.5)^{2} = 5 \times 0.25 = 1.25 m$
The third water drop will be at a height of
$5 - 1.25 = 3.75 m$

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