Question

Water drops fall from a tap on the floor, \(5\) m below, at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from the ground, at the instant when the first drop strikes the ground, is ________ m. (\( g = 10 \, \text{m s}^{-2} \))

• A. \(4.0\)
• B. \(3.8\)
• C. \(4.2\)
• D. \(2.5\)

Hint:

When objects are released at equal intervals, first determine the interval using one complete motion, then analyze the partial motion of the required object.

Answer 1

The Correct Option is C

Concept: Drops are released at equal time intervals. The motion of each drop is uniformly accelerated motion under gravity, starting from rest. The kinematic equation \[ s=\frac{1}{2}gt^2 \] is used.
Step 1: Time taken by the first drop to reach the ground Height of fall \( =5 \) m: \[ 5=\frac{1}{2}\cdot 10 \cdot t^2 \Rightarrow t^2=1 \Rightarrow t=1\text{ s} \] Thus, the first drop hits the ground at \( t=1 \) s.
Step 2: Find the time interval between successive drops The sixth drop begins to fall when the first drop hits the ground. Number of intervals between 1st and 6th drops \( =5 \). Let the time interval between drops be \( \tau \). \[ 5\tau =1 \Rightarrow \tau=0.2\text{ s} \]
Step 3: Time for which the fourth drop has been falling The fourth drop is released at: \[ 3\tau = 3\times0.2 = 0.6\text{ s} \] At the instant the first drop hits the ground (\(t=1\) s), time of fall of the fourth drop: \[ t_4 = 1-0.6 = 0.4\text{ s} \]
Step 4: Distance fallen by the fourth drop \[ s_4=\frac{1}{2}gt_4^2 =\frac{1}{2}\cdot10\cdot(0.4)^2 =0.8\text{ m} \]
Step 5: Height of the fourth drop above the ground \[ h = 5-0.8 = 4.2\text{ m} \]