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

To solve this problem, we need to understand the sequence of falling balls and calculate the height of the \(4^{th}\) ball at a particular moment. Let's go step-by-step:

1. The problem states that balls are dropped from a height of 5 m at regular intervals.
2. The first ball touches the ground when the \(6^{th}\) ball is about to be dropped. This means the time taken for the first ball to travel 5 m is equal to the time interval between dropping successive balls.
3. To calculate the time taken for a ball to drop from 5 m, we use the formula for the time of free fall under gravity, \( t = \sqrt{\frac{2h}{g}} \), where \( h = 5 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity).

\(t = \sqrt{\frac{2 \times 5}{9.8}} = \sqrt{\frac{10}{9.8}}\)

1. Calculate \( t \):

\(t = \sqrt{\frac{10}{9.8}} \approx 1.01 \, \text{s}\)

1. This time (1.01 s) is the interval between dropping successive balls. Since the \(6^{th}\) ball is about to be dropped when the first ball touches the ground, the time elapsed when the \(4^{th}\) ball is in the air is \(3 \times 1.01\).
2. Now, calculate the distance fallen by the \(4^{th}\) ball using the equation: \(h = \frac{1}{2}gt^2\)

\(h = \frac{1}{2} \times 9.8 \times (3 \times 1.01)^2\)

1. Substitute \(t\) and solve for \(h\):

\(h = \frac{1}{2} \times 9.8 \times (3.03)^2 = \frac{1}{2} \times 9.8 \times 9.1809 \approx 45.09 / 2 = 22.545 \, \text{m}\)

1. The \(4^{th}\) ball has fallen approximately 0.8 m at this time, so its height above the ground is:

\((5 - 0.8) \, \text{m} = 4.2 \, \text{m}\)

Therefore, the height of the \(4^{th}\) ball above the ground at the moment when the first ball touches the ground is 4.2 m.