The Rhythm of Falling Drops
Have you ever watched a leaky faucet or a showerhead and noticed the mesmerizing, rhythmic pattern of the falling drops? This problem takes that everyday observation and turns it into a beautiful exercise in kinematics.
The core trick to solving this problem isn't a complex formula; it's understanding the timeline of events. The problem states that the drops fall at regular intervals. When the first drop hits the floor, the third drop is just starting its journey.
Mapping the Time Intervals
Imagine a stopwatch that clicks every time a drop leaves the nozzle. Let's call this interval Δt.
When drop 1 leaves, the time is t=0.
When drop 2 leaves, the time is t=Δt.
When drop 3 leaves, the time is t=2Δt.
The problem tells us that at t=2Δt, the first drop hits the ground. This means the total time of flight for the first drop, let's call it t1, is exactly 2Δt.
At this exact moment, how long has the second drop been falling? It started at t=Δt, so it has been falling for a duration of t2=Δt.
This gives us our master relationship:
t2=2t1
The Master Equation
Now, we can use the second equation of motion to find the total time t1. Since the drops fall from rest, the initial velocity u=0.
Substituting our values for the first drop:
Notice how elegantly the 9.8 cancels out on both sides!
Finding the Second Drop
Since the second drop has been falling for half the time, its time of flight is:
Let's find out how far it has fallen from the nozzle. We'll call this distance x.
The Final Trap
Here is where many students lose marks. They see 2.45 m, spot it in the options, and confidently tick it. But wait! The question specifically asks for the position of the second drop from the floor, not from the nozzle.
To find the height from the floor, we subtract the distance fallen from the total height:
And there we have it! By carefully mapping the timeline and avoiding the final trap, we arrive at the correct answer.