A ball is dropped from a height of `45 m` from the ground. The coefficient of restitution between the ball and the ground is `2/3`. What is the distance travelled by the ball in `4th` second of its motion. Assume negligible time is spent in rebounding. Let `g= 10 ms^(2)`

Options

A. `5m`
B. `20m`
C. `15m`
D. `10m`

Explanation

To solve the problem step by step, we will follow the outlined approach:

### Step 1: Determine the time taken for the ball to reach the ground

The ball is dropped from a height of 45 m. We can use the equation of motion to find the time taken to reach the ground:

\[ s = ut + \frac{1}{2} g t^2 \]

Where:
- \( s = 45 \, \text{m} \) (height)
- \( u = 0 \, \text{m/s} \) (initial velocity, since the ball is dropped)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting the values into the equation:

\[ 45 = 0 \cdot t + \frac{1}{2} \cdot 10 \cdot t^2 \]

This simplifies to:

\[ 45 = 5t^2 \]

Solving for \( t^2 \):

\[ t^2 = \frac{45}{5} = 9 \]

Taking the square root:

\[ t = 3 \, \text{s} \]

### Step 2: Calculate the velocity of the ball just before it hits the ground

Using the formula for velocity just before impact:

\[ v = \sqrt{2gh} \]

Substituting the values:

\[ v = \sqrt{2 \cdot 10 \cdot 45} = \sqrt{900} = 30 \, \text{m/s} \]

### Step 3: Calculate the velocity of the ball after rebounding

The coefficient of restitution \( e \) is given as \( \frac{2}{3} \). The velocity after rebounding is given by:

\[ v' = e \cdot v \]

Substituting the values:

\[ v' = \frac{2}{3} \cdot 30 = 20 \, \text{m/s} \]

### Step 4: Calculate the distance travelled in the 4th second of motion

We need to find the distance travelled during the 4th second. The ball falls for the first 3 seconds and then rebounds.

For the 4th second, we need to calculate the distance travelled during the 1st second of the rebound.

Using the equation of motion again for the rebound:

\[ s' = v' t - \frac{1}{2} g t^2 \]

Where:
- \( v' = 20 \, \text{m/s} \) (velocity after rebound)
- \( g = 10 \, \text{m/s}^2 \)
- \( t = 1 \, \text{s} \) (time during the 4th second)

Substituting the values:

\[ s' = 20 \cdot 1 - \frac{1}{2} \cdot 10 \cdot (1)^2 \]

This simplifies to:

\[ s' = 20 - 5 = 15 \, \text{m} \]

### Final Answer

The distance travelled by the ball in the 4th second of its motion is **15 m**.

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