Question

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. 2.45 m
• B. 7.35 m
• C. 2.94 m
• D. 4.18 m

Hint:

In problems involving objects dropped at regular intervals, the ratio of distances covered by consecutive drops from the top follows the pattern \(1:4:9: \dots\) for equal time intervals. Here, the first drop has fallen for time \(2\Delta t\) and the second for \( \Delta t \). The ratio of distances fallen is \( (2\Delta t)^2 : (\Delta t)^2 = 4:1 \). So, \(h_2 = H/4 = 9.8/4 = 2.45\) m from the top. Position from floor is \( H - H/4 = 3H/4 = 3 \times 2.45 = 7.35 \) m.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a scenario where water drops fall from a height of 9.8 m at regular time intervals. We need to find the position of the second drop when the first drop hits the floor, given that the third drop starts to fall at that same instant.
Step 2: Key Formula or Approach:
We will use the equation of motion for an object in free fall:
\[ s = ut + \frac{1}{2}at^2 \] Since the drops start from rest, their initial velocity \( u = 0 \). The acceleration is due to gravity, \( a = g \). The formula becomes:
\[ h = \frac{1}{2}gt^2 \] where \( h \) is the distance fallen in time \( t \). We are given \( g = 9.8 \, \text{m/s}^2 \).
Step 3: Detailed Explanation:
First, let's calculate the total time taken by the first drop to reach the floor.
Given height \( H = 9.8 \) m.
\[ H = \frac{1}{2}gt^2_{\text{total}} \] \[ 9.8 = \frac{1}{2} \times 9.8 \times t^2_{\text{total}} \] \[ t^2_{\text{total}} = 2 \] \[ t_{\text{total}} = \sqrt{2} \, \text{s} \] The drops fall at a regular interval. Let this interval be \( \Delta t \).
The first drop starts at \( t = 0 \).
The second drop starts at \( t = \Delta t \).
The third drop starts at \( t = 2\Delta t \).
The problem states that when the first drop strikes the floor (at \( t = t_{\text{total}} \)), the third drop begins to fall.
This means \( t_{\text{total}} = 2\Delta t \).
So, \( \sqrt{2} = 2\Delta t \), which gives \( \Delta t = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \, \text{s} \).
Now, we need to find the position of the second drop when the first drop hits the floor, i.e., at time \( t = t_{\text{total}} = \sqrt{2} \) s.
The second drop started falling at \( t = \Delta t \). So, the time for which the second drop has been falling is:
\[ t_{\text{2nd drop}} = t_{\text{total}} - \Delta t = 2\Delta t - \Delta t = \Delta t = \frac{1}{\sqrt{2}} \, \text{s} \] The distance fallen by the second drop in this time is:
\[ h_2 = \frac{1}{2}gt^2_{\text{2nd drop}} \] \[ h_2 = \frac{1}{2} \times 9.8 \times \left(\frac{1}{\sqrt{2}}\right)^2 \] \[ h_2 = \frac{1}{2} \times 9.8 \times \frac{1}{2} = \frac{9.8}{4} = 2.45 \, \text{m} \] This is the distance from the nozzle (the top). The question asks for the position from the floor.
Position from floor = Total Height - Distance fallen
\[ \text{Position}_2 = H - h_2 = 9.8 - 2.45 = 7.35 \, \text{m} \] Step 4: Final Answer:
The position of the second drop from the floor is 7.35 m.