Water drops fall from the nozzle of a shower 5 m high on the floor. The drops are released at regular intervals of time such that the first drop reaches the ground when sixth drop is released from the nozzle. Taking g = 10 m/s2. What is the height of the fourth drop from the ground?
4.2 m
Understanding Water Drop Motion from a Shower
This problem involves analyzing the motion of water drops falling from a height under gravity, released at regular time intervals. We are given the total height the drops fall from and the condition that the first drop reaches the ground exactly when the sixth drop is released. We need to find the height of the fourth drop from the ground at that specific moment.
Step-by-Step Solution for Falling Water Drops
1. Calculate the Total Time for One Drop to Fall
The water drops fall from a height of 5 m. We can use the equation of motion for an object under free fall:
\(s = ut + \frac{1}{2}gt^2\)
Where:
- \(s\) is the displacement (5 m)
- \(u\) is the initial velocity (0 m/s, as the drops are released)
- \(g\) is the acceleration due to gravity (10 m/s²)
- \(t\) is the time taken
Substituting the values:
\(5 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2\)
\(5 = 5t^2\)
\(t^2 = \frac{5}{5} = 1\)
\(t = \sqrt{1} = 1 \text{ second}\)
So, it takes 1 second for a single drop to fall from the nozzle to the ground.
2. Determine the Time Interval Between Drops
The problem states that the first drop reaches the ground when the sixth drop is released. This means that the total time of fall for the first drop (1 second) is spread over the intervals between the release of 6 drops.
Let the regular time interval between the release of successive drops be \(\Delta t\).
Consider the sequence of events:
- Drop 1 released
- Drop 2 released at time \(\Delta t\) after Drop 1
- Drop 3 released at time \(2\Delta t\) after Drop 1
- Drop 4 released at time \(3\Delta t\) after Drop 1
- Drop 5 released at time \(4\Delta t\) after Drop 1
- Drop 6 released at time \(5\Delta t\) after Drop 1 (At this moment, Drop 1 hits the ground)
The total time elapsed from the release of the first drop to the release of the sixth drop is \(5\Delta t\). This total time is equal to the time it takes for the first drop to hit the ground, which we calculated as 1 second.
Therefore, \(5\Delta t = 1\) second.
\(\Delta t = \frac{1}{5} = 0.2 \text{ seconds}\)
The time interval between the release of each water drop is 0.2 seconds.
3. Calculate the Time of Fall for the Fourth Drop
We want to find the height of the fourth drop from the ground when the first drop hits the ground (and the sixth drop is released).
At this moment (1 second after the first drop was released):
- Drop 1 has been falling for 1.0 seconds (hit the ground).
- Drop 2 has been falling for \(1.0 - 0.2 = 0.8\) seconds.
- Drop 3 has been falling for \(1.0 - 2 \times 0.2 = 1.0 - 0.4 = 0.6\) seconds.
- Drop 4 has been falling for \(1.0 - 3 \times 0.2 = 1.0 - 0.6 = 0.4\) seconds.
- Drop 5 has been falling for \(1.0 - 4 \times 0.2 = 1.0 - 0.8 = 0.2\) seconds.
- Drop 6 has been falling for 0 seconds (just released).
The fourth drop has been falling for 0.4 seconds.
4. Calculate the Distance Fallen by the Fourth Drop
Now, we calculate how far the fourth drop has fallen in 0.4 seconds using the same equation of motion:
\(s_4 = ut_4 + \frac{1}{2}gt_4^2\)
Where:
- \(u\) is the initial velocity (0 m/s)
- \(g\) is the acceleration due to gravity (10 m/s²)
- \(t_4\) is the time the fourth drop has been falling (0.4 seconds)
Substituting the values:
\(s_4 = 0 \cdot (0.4) + \frac{1}{2} \cdot 10 \cdot (0.4)^2\)
\(s_4 = 5 \cdot (0.16)\)
\(s_4 = 0.8 \text{ meters}\)
The fourth drop has fallen 0.8 meters from the nozzle.
5. Calculate the Height of the Fourth Drop from the Ground
The total height of the shower nozzle from the ground is 5 m. The fourth drop has fallen 0.8 m from the top.
Height of the fourth drop from the ground = Total height - Distance fallen by the fourth drop
Height \(h_4 = 5 \text{ m} - 0.8 \text{ m} = 4.2 \text{ m}\)
Therefore, the height of the fourth drop from the ground when the first drop reaches the ground is 4.2 meters.
| Drop Number | Time Elapsed Since Release (s) | Distance Fallen (m) | Height from Ground (m) |
|---|---|---|---|
| 1st | 1.0 | \(0 + \frac{1}{2} \cdot 10 \cdot (1.0)^2 = 5.0\) | \(5 - 5.0 = 0.0\) |
| 2nd | 0.8 | \(0 + \frac{1}{2} \cdot 10 \cdot (0.8)^2 = 3.2\) | \(5 - 3.2 = 1.8\) |
| 3rd | 0.6 | \(0 + \frac{1}{2} \cdot 10 \cdot (0.6)^2 = 1.8\) | \(5 - 1.8 = 3.2\) |
| 4th | 0.4 | \(0 + \frac{1}{2} \cdot 10 \cdot (0.4)^2 = 0.8\) | \(5 - 0.8 = 4.2\) |
| 5th | 0.2 | \(0 + \frac{1}{2} \cdot 10 \cdot (0.2)^2 = 0.2\) | \(5 - 0.2 = 4.8\) |
| 6th | 0.0 | \(0 + \frac{1}{2} \cdot 10 \cdot (0.0)^2 = 0.0\) | \(5 - 0.0 = 5.0\) |
The calculation confirms that the height of the fourth drop from the ground is 4.2 m.
Revision Table: Key Concepts
| Concept | Description | Relevant Formula |
|---|---|---|
| Free Fall | Motion under the influence of gravity only, assuming no air resistance. | Various kinematic equations |
| Acceleration due to Gravity (g) | Constant acceleration experienced by objects in free fall near the Earth's surface. (Approx. 9.8 m/s², given as 10 m/s² in this problem) | - |
| Kinematic Equation for Displacement | Relates displacement, initial velocity, time, and acceleration. | \(s = ut + \frac{1}{2}at^2\) |
| Regular Time Intervals | Events occurring with a constant duration between them. Important for timing multiple objects in motion. | \(\Delta t\) |
Additional Information: Motion Under Gravity
Problems involving objects in free fall or projected vertically are common in physics. The key is to apply the appropriate kinematic equations. For motion under gravity, the acceleration \(a\) is equal to \(g\) (acceleration due to gravity), which is usually taken as 9.8 m/s² or rounded to 10 m/s² for simpler calculations, as done in this problem. The direction of velocity and displacement must be considered when using these equations (upward usually negative, downward usually positive, or vice versa, consistently).
In this problem, since all drops are falling downwards, we take downward as positive. The initial velocity (\(u\)) for drops released from rest is 0. The time intervals between drops create a sequence where each drop has been falling for a different duration at any given moment. By calculating the time elapsed for a specific drop, we can determine its position.
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