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, we will break it down into parts:

### Step 1: Calculate the velocity of the ball just before it enters the water.

The velocity \( V \) of the ball just before it enters the water can be calculated using the formula for free fall:

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

Where:
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 19.6 \, \text{m} \) (height from which the ball falls)

Substituting the values:

\[ V = \sqrt{2 \times 9.8 \times 19.6} \]
\[ V = \sqrt{384.16} \approx 19.6 \, \text{m/s} \]

### Step 2: Determine the density of the ball and the water.

Let the density of water be \( \rho_w \) and the density of the ball be \( \rho_b \). Given that the density of the ball is half that of water:

\[ \rho_b = \frac{1}{2} \rho_w \]

### Step 3: Calculate the upthrust (buoyant force) acting on the ball when it is submerged.

The upthrust \( F_b \) can be calculated using Archimedes' principle:

\[ F_b = \text{Volume of ball} \times \text{Density of water} \times g \]

Let the volume of the ball be \( V \).

\[ F_b = V \cdot \rho_w \cdot g \]

### Step 4: Calculate the weight of the ball.

The weight \( W \) of the ball is given by:

\[ W = \text{Volume of ball} \times \text{Density of ball} \times g \]
\[ W = V \cdot \rho_b \cdot g = V \cdot \left(\frac{1}{2} \rho_w\right) \cdot g \]

### Step 5: Determine the net force acting on the ball when submerged.

The net force \( F_{net} \) acting on the ball when it is submerged in water is given by:

\[ F_{net} = F_b - W \]
\[ F_{net} = V \cdot \rho_w \cdot g - V \cdot \left(\frac{1}{2} \rho_w\right) \cdot g \]
\[ F_{net} = V \cdot g \cdot \left(\rho_w - \frac{1}{2} \rho_w\right) = V \cdot g \cdot \frac{1}{2} \rho_w \]

### Step 6: Calculate the acceleration of the ball when submerged.

The mass \( m \) of the ball is:

\[ m = V \cdot \rho_b = V \cdot \left(\frac{1}{2} \rho_w\right) \]

Using Newton's second law, the acceleration \( a \) of the ball when submerged is:

\[ a = \frac{F_{net}}{m} = \frac{V \cdot g \cdot \frac{1}{2} \rho_w}{V \cdot \left(\frac{1}{2} \rho_w\right)} = g \]

Thus, the acceleration \( a = 9.8 \, \text{m/s}^2 \).

### Step 7: Calculate the depth to which the ball will sink.

Using the equation of motion, we can find the distance \( d \) the ball sinks in water:

\[ d = \frac{V^2}{2a} \]

Substituting the values:

\[ d = \frac{(19.6)^2}{2 \times 9.8} = \frac{384.16}{19.6} = 19.6 \, \text{m} \]

### Step 8: Calculate the time taken for the ball to return to the surface.

The time \( T \) taken to return to the surface can be calculated using:

\[ T = \frac{2V}{a} \]

Substituting the values:

\[ T = \frac{2 \times 19.6}{9.8} = 4 \, \text{s} \]

### Final Answers:

1. The depth to which the ball will sink is **19.6 m**.
2. The time taken to return to the water surface is **4 seconds**.