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 the ground, at the instant when the first drop strikes the ground, is ________ m. (\( g = 10 \, \text{m s}^{-2} \))

• A. \(4.0\)
• B. \(3.8\)
• C. \(4.2\)
• D. \(2.5\)

Hint:

When objects are released at equal intervals, first determine the interval using one complete motion, then analyze the partial motion of the required object.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a motion under gravity problem with multiple objects released at equal time intervals. We need to determine the position of a specific drop when another hits the ground.
Step 2: Key Formula or Approach:
Distance fallen: \(h = \frac{1}{2}gt^2\).
Height from ground: \(H = 5 - h\).
Step 3: Detailed Explanation:
Total height \(H = 5 \text{ m}\).
Time for the 1st drop to hit the ground:
\[ t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \text{ s} \]
At \(t = 1 \text{ s}\), the 6th drop begins to fall. This means 5 equal time intervals (\(\Delta t\)) have occurred since the 1st drop fell.
\[ 5\Delta t = 1 \text{ s} \Rightarrow \Delta t = 0.2 \text{ s} \]
At the instant the 1st drop hits the ground (\(t = 1 \text{ s}\)):
- 6th drop has been falling for \(0 \text{ s}\).
- 5th drop has been falling for \(1\Delta t = 0.2 \text{ s}\).
- 4th drop has been falling for \(2\Delta t = 0.4 \text{ s}\).
Distance fallen by the 4th drop:
\[ h_4 = \frac{1}{2} g (0.4)^2 = \frac{1}{2} \times 10 \times 0.16 = 0.8 \text{ m} \]
Height from the ground:
\[ H_4 = 5 - 0.8 = 4.2 \text{ m} \]
Step 4: Final Answer:
The height of the fourth drop from the ground is \(4.2 \text{ m}\).