An open tank containing non-viscous liquid to a height of 5 m is placed over the ground. A heavy sperical ball falls from height 40 m over the ground which ball will go back. Collision between ball and bottom of tank is perfectly elastic

Options

A. 45 m
B. 35m
C. 40 m
D. 20 m

Explanation

To solve the problem, we need to analyze the situation step by step:

### Step 1: Understand the setup
We have an open tank filled with a non-viscous liquid (water) to a height of 5 meters. A heavy spherical ball is dropped from a height of 40 meters above the ground. The collision between the ball and the bottom of the tank is perfectly elastic.

**Hint:** Visualize the setup by sketching the tank and the ball's trajectory.

### Step 2: Determine the initial energy of the ball
When the ball is dropped from a height of 40 meters, it possesses gravitational potential energy given by the formula:
\[ PE = mgh \]
where:
- \( m \) = mass of the ball
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (40 m)

**Hint:** Remember that potential energy converts to kinetic energy as the ball falls.

### Step 3: Calculate the velocity of the ball just before impact
As the ball falls, it converts its potential energy into kinetic energy. Just before hitting the water surface, all potential energy will have converted into kinetic energy:
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to the kinetic energy at the moment before impact:
\[ mgh = \frac{1}{2} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh = \frac{1}{2} v^2 \]
Solving for \( v \):
\[ v = \sqrt{2gh} = \sqrt{2 \cdot 9.81 \cdot 40} \]

**Hint:** Use the values of \( g \) and \( h \) to compute the velocity.

### Step 4: Analyze the collision with the water
When the ball hits the water, it will experience a change in medium, but since the collision with the bottom of the tank is perfectly elastic, it will not lose any energy during the collision.

**Hint:** Recall that in a perfectly elastic collision, the ball retains its kinetic energy.

### Step 5: Determine the height the ball will reach after the collision
Since the collision is perfectly elastic, the ball will rebound with the same kinetic energy it had just before hitting the bottom of the tank. Therefore, it will rise back to the same height from which it fell (40 m).

**Hint:** Use the conservation of energy principle to confirm that the ball will reach the same height.

### Conclusion
The ball will go back to a height of 40 meters after the collision with the bottom of the tank.

### Final Answer
The ball will rise back to a height of **40 meters**.