A ball is thrown vertically downwards from a height of 20 m with an initial velocity `upsilon_0` It collides with the ground loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity `upsilon_0` is (Take ` g = 10ms^(-2)` )

Options

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

Explanation

To solve the problem, we need to find the initial velocity \( v_0 \) of a ball thrown vertically downwards from a height of 20 m, which loses 50% of its energy upon colliding with the ground and rebounds to the same height.

### Step 1: Calculate the velocity just before impact with the ground

Using the third equation of motion:
\[ v^2 = v_0^2 + 2gh \]
where:
- \( v \) is the final velocity just before impact,
- \( v_0 \) is the initial velocity,
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 20 \, \text{m} \) (height).

Rearranging the equation gives:
\[ v^2 = v_0^2 + 2 \cdot 10 \cdot 20 \]
\[ v^2 = v_0^2 + 400 \]

### Step 2: Calculate the kinetic energy just before impact

The kinetic energy (KE) just before the collision is given by:
\[ \text{KE} = \frac{1}{2} mv^2 \]

### Step 3: Calculate the kinetic energy after the collision

Since the ball loses 50% of its energy during the collision, the kinetic energy after the collision is:
\[ \text{KE}_{\text{after}} = \frac{1}{2} \text{KE} = \frac{1}{2} \left( \frac{1}{2} mv^2 \right) = \frac{1}{4} mv^2 \]

### Step 4: Relate the kinetic energy after the collision to the velocity after the collision

Let \( v_1 \) be the velocity after the collision. The kinetic energy after the collision can also be expressed as:
\[ \text{KE}_{\text{after}} = \frac{1}{2} mv_1^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ \frac{1}{4} mv^2 = \frac{1}{2} mv_1^2 \]
Cancelling \( m \) from both sides and simplifying gives:
\[ \frac{1}{4} v^2 = \frac{1}{2} v_1^2 \]
\[ v_1^2 = \frac{1}{2} v^2 \]

### Step 5: Calculate the height reached after the rebound

The ball rebounds to the same height \( h \) after the collision. Using the equation of motion again:
\[ 0 = v_1^2 - 2gh \]
This gives:
\[ v_1^2 = 2gh \]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( h = 20 \, \text{m} \):
\[ v_1^2 = 2 \cdot 10 \cdot 20 = 400 \]

### Step 6: Relate \( v_1^2 \) to \( v_0^2 \)

From the previous step, we have:
\[ v_1^2 = \frac{1}{2} v^2 \]
Substituting \( v^2 \) from Step 1:
\[ 400 = \frac{1}{2} (v_0^2 + 400) \]
Multiplying through by 2 gives:
\[ 800 = v_0^2 + 400 \]
Rearranging gives:
\[ v_0^2 = 800 - 400 = 400 \]

### Step 7: Calculate \( v_0 \)

Taking the square root:
\[ v_0 = \sqrt{400} = 20 \, \text{m/s} \]

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

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