A ball after falling a distance of 5 meter rest hits elastically the floor of a lift and rebounds. At the time of impact the lift was moving up with a velocity of 1m/sec. The velocity with which the ball rebounds just after impact is : `(g=10m//sec^(2))`

Options

A. 10 m/sec
B. 11m/sec
C. 12m/sec
D. 13m/sec

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the velocity of the ball just before impact.
The ball falls from rest under the influence of gravity. We can use the equation of motion to find the velocity just before it hits the floor:

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

Where:
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 5 \, \text{m} \) (height from which the ball falls)

Substituting the values:

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

### Step 2: Identify the velocities at the time of impact.
At the time of impact:
- The velocity of the ball \( v_B = 10 \, \text{m/s} \) (downward)
- The velocity of the lift \( v_L = 1 \, \text{m/s} \) (upward)

### Step 3: Use the principle of conservation of momentum for elastic collision.
In an elastic collision, the velocity of the ball after the collision can be calculated using the formula:

\[ v' = v_B + 2v_L \]

Where:
- \( v' \) is the velocity of the ball after the impact.
- \( v_B \) is the velocity of the ball just before impact.
- \( v_L \) is the velocity of the lift.

### Step 4: Substitute the values into the formula.
Now substituting the values we have:

\[ v' = 10 \, \text{m/s} + 2 \times 1 \, \text{m/s} \]
\[ v' = 10 \, \text{m/s} + 2 \, \text{m/s} = 12 \, \text{m/s} \]

### Conclusion:
The velocity with which the ball rebounds just after impact is \( 12 \, \text{m/s} \).

---