A ball is thrown vertically downwards from a height of 20m with an intial velocity `v_(0)`. It collides with the ground, loses 50% of its energy in collision and rebounds to the same height. The intial velocity `v_(0)` is (Take, g =10 `ms^(-2)`)

Options

A. `14ms^(-1)`
B. `20ms^(-1)`
C. `28ms^(-1)`
D. `10ms^(-1)`

Explanation

To find the initial velocity \( v_0 \) of the ball thrown vertically downwards from a height of 20 m, we can follow these steps:

### Step 1: Calculate the potential energy at the height of 20 m
The potential energy (PE) at height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m \) is the mass of the ball,
- \( g \) is the acceleration due to gravity (10 m/s²),
- \( h \) is the height (20 m).

Substituting the values:
\[ PE = mg \cdot 20 \]

### Step 2: Calculate the kinetic energy just before the collision
When the ball is thrown downwards with an initial velocity \( v_0 \), its total mechanical energy just before hitting the ground is the sum of its potential energy and kinetic energy (KE):
\[ KE = \frac{1}{2} mv_0^2 \]
At the moment before collision, the potential energy will be zero (as height is zero), so:
\[ \text{Total Energy} = KE + PE = \frac{1}{2} mv_0^2 + mg \cdot 20 \]

### Step 3: Calculate the energy after the collision
After the collision, the ball loses 50% of its energy. Therefore, the energy after the collision is:
\[ \text{Energy after collision} = \frac{1}{2} \left( \frac{1}{2} mv_0^2 + mg \cdot 20 \right) \]

### Step 4: Set the energy after the collision equal to the potential energy at the rebound height
The ball rebounds to the same height of 20 m, so the potential energy after the collision is:
\[ PE' = mg \cdot 20 \]
Setting the energy after the collision equal to the potential energy at the rebound height:
\[ \frac{1}{2} \left( \frac{1}{2} mv_0^2 + mg \cdot 20 \right) = mg \cdot 20 \]

### Step 5: Simplify the equation
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} \left( \frac{1}{2} v_0^2 + 20g \right) = 20g \]
Multiplying through by 2:
\[ \frac{1}{2} v_0^2 + 20g = 40g \]
Subtracting \( 20g \) from both sides:
\[ \frac{1}{2} v_0^2 = 20g \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ \frac{1}{2} v_0^2 = 20 \cdot 10 \]
\[ \frac{1}{2} v_0^2 = 200 \]
Multiplying both sides by 2:
\[ v_0^2 = 400 \]
Taking the square root:
\[ v_0 = 20 \, \text{m/s} \]

### Final Answer
The initial velocity \( v_0 \) is \( 20 \, \text{m/s} \).

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