Q. Water drops fall from a tap on the floor 5m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be, from ground, at the instant when first drop strikes the ground, will be ($g =$10$ms^{- 2}$)
A) 1.25 m
B) 2.15 m
C) 2.73 m
D) 3.75 m
Explanation
(d) By the time 5$^{th}$ water drop starts falling, the first water drop reaches the ground.
As $u = 0,$ $h = \frac{1}{2}gt^{2} = \frac{1}{2} \times 10 \times t^{2}$
or $5 = \frac{1}{2} \times 10 \times t^{2}$ or $t = 1$ s
Hence, the interval of each water drop$\ = \frac{1s}{4} = 0.25$s.
When the 5$^{th}$ drop starts its journey towards ground, the third drop travels in air for
$t_{1} = 0.25 + 0.25 = 0.5s$
$\therefore\ $ Height (distance) covered by 3$^{rd}$ drop in air is
$h_{1} = \frac{1}{2}gt_{1}^{2} = \frac{1}{2} \times 10 \times (0.5)^{2}$
$= 5 \times 0.25 = 1.25$m
So, third water drop will be at a height of
$= 5 - 1.25 = 3.75$m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.