Question

Balls are dropped at regular intervals from height 5 m. If the first ball touches the ground when \(6^{th}\) ball is about to be dropped, find the height of \(4^{th}\) ball above the ground at the same instant :

• A. 4.1 m
• B. 4.2 m
• C. 4.3 m
• D. 4.4 m
• E. None of these

Hint:

For \(N\) balls, there are \(N-1\) intervals. Calculate the interval \(\Delta t\), then the time fallen by the \(k^{th}\) ball is \((N-k)\Delta t\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The total time of flight for the first ball is divided into equal intervals between the subsequent balls being dropped.
Step 2: Key Formula or Approach:
Time of flight: \(T = \sqrt{\frac{2H}{g}}\).
Distance fallen: \(s = \frac{1}{2}gt^2\).
Step 3: Detailed Explanation:
Height \(H = 5 \text{ m}\). Let \(g = 10 \text{ m/s}^2\).
Time for the first ball to hit ground: \(T = \sqrt{\frac{2 \times 5}{10}} = 1 \text{ s}\).
When the \(6^{th}\) ball is dropped, the \(1^{st}\) ball has hit the ground. There are 5 intervals in 1 second.
Time interval between drops \(\Delta t = \frac{1}{5} = 0.2 \text{ s}\).
At the instant \(6^{th}\) ball is dropped:
The \(4^{th}\) ball has been falling for a time \(t_4 = (6-4)\Delta t = 2 \times 0.2 = 0.4 \text{ s}\).
Distance fallen by \(4^{th}\) ball:
\[ s_4 = \frac{1}{2} g t_4^2 = \frac{1}{2} (10) (0.4)^2 = 5 \times 0.16 = 0.8 \text{ m} \]
Height above ground:
\[ h = H - s_4 = 5 - 0.8 = 4.2 \text{ m} \]
Step 4: Final Answer:
The height of the 4th ball above the ground is 4.2 m.