A wooden ball of density D is immersed in water of density d to a depth h//2 below the surface of water and then relased. To what height will the ball jump out of water ?

Explanation

To solve the problem of how high a wooden ball of density \( D \) will jump out of water after being released from a depth of \( \frac{h}{2} \), we can follow these steps:

### Step 1: Understand the Forces Acting on the Ball
When the ball is submerged in water, two main forces act on it:
1. The gravitational force (weight) acting downward:
\[ F_g = Mg = V \cdot D \cdot g \]
where \( V \) is the volume of the wooden ball, \( D \) is the density of the wooden ball, and \( g \) is the acceleration due to gravity.

2. The buoyant force acting upward (upthrust):
\[ F_b = V \cdot d \cdot g \]
where \( d \) is the density of water.

### Step 2: Calculate the Net Force
The net force acting on the ball when it is released can be calculated as:
\[ F_{net} = F_b - F_g = V \cdot d \cdot g - V \cdot D \cdot g = Vg(d - D) \]

### Step 3: Determine the Acceleration
Using Newton's second law, we can find the acceleration \( a \) of the ball:
\[ F_{net} = M \cdot a \implies Vg(d - D) = V \cdot D \cdot a \]
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ g(d - D) = D \cdot a \implies a = \frac{g(d - D)}{D} \]

### Step 4: Calculate the Velocity at the Water Surface
The ball is released from a depth of \( \frac{h}{2} \) and accelerates upwards. We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where \( u = 0 \) (initial velocity), \( s = \frac{h}{2} \) (distance moved), and \( a \) is the acceleration we found:
\[ v^2 = 0 + 2 \cdot \frac{g(d - D)}{D} \cdot \frac{h}{2} \]
Thus,
\[ v^2 = \frac{g(d - D)h}{D} \]

### Step 5: Calculate the Maximum Height Above Water Surface
After reaching the water surface, the ball will continue to move upwards until its velocity becomes zero. Using the same kinematic equation:
\[ 0 = v^2 - 2gH \]
where \( H \) is the height above the water surface:
\[ 0 = \frac{g(d - D)h}{D} - 2gH \]
Rearranging gives:
\[ 2gH = \frac{g(d - D)h}{D} \implies H = \frac{(d - D)h}{2D} \]

### Final Result
Thus, the height \( H \) to which the ball will jump out of the water is:
\[ H = \frac{(d - D)h}{2D} \]