A ball is dropped from a height of 5 m, if it rebound upto height of 1.8 m, then the ratio of velocities of the ball after and before rebound is :

Options

A. `3/5`
B. `2/5`
C. `1/5`
D. `4/5`

Explanation

To find the ratio of the velocities of the ball after and before the rebound, we can follow these steps:

### Step 1: Determine the velocity just before the ball hits the ground.
The ball is dropped from a height of 5 meters. We can use the third equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity just before hitting the ground
- \( u \) = initial velocity (which is 0 since the ball is dropped)
- \( a \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( s \) = distance fallen (5 m)

Substituting the values:

\[ v^2 = 0 + 2 \times 10 \times 5 \]
\[ v^2 = 100 \]
\[ v = \sqrt{100} = 10 \, \text{m/s} \]

### Step 2: Determine the velocity just after the ball rebounds.
The ball rebounds to a height of 1.8 meters. At the highest point of the rebound, the final velocity is 0. We again use the third equation of motion:

\[ v^2 = u^2 + 2as \]

Where:
- \( v \) = final velocity (0 at the highest point)
- \( u \) = initial velocity after rebound (which we need to find)
- \( a \) = acceleration due to gravity (which is negative, so \( -10 \, \text{m/s}^2 \))
- \( s \) = height reached (1.8 m)

Substituting the values:

\[ 0 = u^2 + 2 \times (-10) \times 1.8 \]
\[ 0 = u^2 - 36 \]
\[ u^2 = 36 \]
\[ u = \sqrt{36} = 6 \, \text{m/s} \]

### Step 3: Calculate the ratio of the velocities after and before the rebound.
Now we have:
- Velocity before the rebound \( v = 10 \, \text{m/s} \)
- Velocity after the rebound \( u = 6 \, \text{m/s} \)

The ratio of the velocities after and before the rebound is:

\[ \text{Ratio} = \frac{u}{v} = \frac{6}{10} = \frac{3}{5} \]

### Final Answer:
The ratio of the velocities of the ball after and before the rebound is \( \frac{3}{5} \).

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