A soild ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. Upto what depth will the ball go. How much time will it take to come again to the water surface? Neglect air resistandce and viscosity effects in water. (Take `g=9.8 m//S^(2))`.

Explanation

To solve the problem step by step, let's break it down into manageable parts:

### Step 1: Understanding the Problem
We have a solid ball with a density that is half that of water. The ball falls freely from a height of 19.6 m and then enters the water. We need to determine how deep the ball will go into the water and how much time it will take to return to the water's surface.

### Step 2: Define the Densities
Let:
- The density of the ball = \( \rho \)
- The density of water = \( 2\rho \)

### Step 3: Calculate the Forces Acting on the Ball in Water
When the ball enters the water, it experiences two main forces:
1. The buoyant force \( F_b \) acting upwards, which is equal to the weight of the water displaced by the ball.
2. The weight of the ball \( W \) acting downwards.

The buoyant force can be calculated as:
\[ F_b = V \cdot \text{density of water} \cdot g = V \cdot (2\rho) \cdot g \]
The weight of the ball is:
\[ W = V \cdot \text{density of ball} \cdot g = V \cdot \rho \cdot g \]

### Step 4: Set Up the Equation of Motion
The net force \( F \) acting on the ball when it is submerged is:
\[ F = F_b - W = V \cdot (2\rho) \cdot g - V \cdot \rho \cdot g \]
\[ F = Vg(2\rho - \rho) = Vg\rho \]

Using Newton's second law \( F = ma \), where \( m \) is the mass of the ball:
\[ ma = Vg\rho \]
Since \( m = V \cdot \rho \), we can substitute:
\[ V\rho a = Vg\rho \]
Dividing both sides by \( V\rho \) (assuming \( V \neq 0 \)):
\[ a = g \]

### Step 5: Calculate the Depth the Ball Will Sink
The ball will decelerate until it comes to rest. The distance \( d \) it sinks can be calculated using the kinematic equation:
\[ v^2 = u^2 + 2ad \]
Where:
- \( v = 0 \) (final velocity when it stops)
- \( u = \text{velocity just before entering water} = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 19.6} = 19.6 \, \text{m/s} \)
- \( a = -g = -9.8 \, \text{m/s}^2 \)

Setting up the equation:
\[ 0 = (19.6)^2 + 2(-9.8)d \]
\[ 0 = 384.16 - 19.6d \]
\[ 19.6d = 384.16 \]
\[ d = \frac{384.16}{19.6} = 19.6 \, \text{m} \]

### Step 6: Calculate the Time to Return to the Surface
The time taken to reach the maximum depth can be calculated using:
\[ t = \frac{v - u}{a} \]
Where \( v = 0 \) (at maximum depth), \( u = 19.6 \, \text{m/s} \), and \( a = -9.8 \, \text{m/s}^2 \):
\[ t = \frac{0 - 19.6}{-9.8} = 2 \, \text{s} \]

Since the ball takes the same time to return to the surface, the total time taken is:
\[ \text{Total time} = 2t = 2 \cdot 2 = 4 \, \text{s} \]

### Final Answers
- The depth to which the ball sinks is \( 19.6 \, \text{m} \).
- The total time taken to return to the water surface is \( 4 \, \text{s} \).