© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Determine the total time for a drop to fall
The drop is released from rest at a height of 9.8 m and falls under gravity. Using the relation
$h = \tfrac{1}{2} g t^2,$
where $h = 9.8\text{ m}$ and $g = 9.8\text{ m/s}^2$, we get:
$9.8 = \tfrac{1}{2} \times 9.8 \times t^2 \quad \Longrightarrow \quad t^2 = 2 \quad \Longrightarrow \quad t = \sqrt{2} \approx 1.414\,\text{s}.$
This $t$ is the total time taken by the first drop to reach the floor.
Step 2: Relate the interval between drops to the total time
According to the problem, when the first drop hits the floor (at $t = 1.414\,\text{s}$), the third drop has just been released. If $\Delta t$ is the time interval between consecutive drops, the first drop starts at $t=0$, the second at $t = \Delta t$, and the third at $t = 2\Delta t.$
Since the third drop starts at $t = 1.414\,\text{s}$, we have
$2\Delta t = 1.414 \quad \Longrightarrow \quad \Delta t = 0.707\,\text{s}.$
Step 3: Find the falling time of the second drop
The second drop begins to fall at $t = \Delta t = 0.707\,\text{s}.$ By the time the first drop strikes the floor ($t = 1.414\,\text{s}$), the second drop would have been falling for
$(1.414 - 0.707) = 0.707\,\text{s}.$
Step 4: Calculate the distance covered by the second drop
In time $0.707\,\text{s}$, starting from rest, the distance $d$ covered by the second drop is:
$d = \tfrac{1}{2} g (0.707)^2 = \tfrac{1}{2} \times 9.8 \times (0.707)^2.$
Note that $(0.707)^2 \approx 0.5$. So the distance covered is:
$d = \tfrac{1}{2} \times 9.8 \times 0.5 = 2.45\,\text{m}.$
This distance is measured from the nozzle (top). Hence, the height of the second drop above the floor is:
$9.8 - 2.45 = 7.35\,\text{m}.$
Step 5: Conclude the position of the second drop
Therefore, when the first drop reaches the floor, the second drop is located
$7.35\,\text{m}$
above the floor.
