Successive Drops at Regular Intervals
If drops are released every tau and the first has fallen for T, the kth released drop has fall time T-(k-1)tau when non-negative. From rest, its fall distance is (1/2)g[T-(k-1)tau]².
Why this shows up in the exam
Leaking-tap spacing · Counting drops in flight · Finding regular release rates
Learn the idea
Regularly released drops share one clock but have different times of fall. At one observation instant, the earliest drop has been falling longest and lies lowest. Each later drop has a fall time shorter by one release interval.
🧠 Memory hook: Same snapshot, staircase of fall times.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- t_k = T-(k-1)tau — fall duration of the kth drop at a common observation time
- s_k = (1/2)g t_k² — distance fallen from rest by the kth drop
How to approach it
- 1Create a release-time timeline
- 2Use one common observation instant
- 3Compute each drop's fall time before its position
Common slip-ups that cost marks
- •Giving every drop the same fall time
- •Confusing release interval with fall time
- •Measuring height from the wrong endpoint
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A particle has initial speed 2 m/s and constant acceleration 3 m/s^2 for 4 s. What distance does it cover?
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