A ball falls under gravity from a height of 10 m with an initial downward velocity u. It collides with the ground, losses 50% of its energy in collision and then rises back to the same height. The initial velocity u is

Options

A. `7 m s^(-1)`
B. `25 m s^(-1)`
C. `14 m s^(-1)`
D. `28 m s^(-1)`

Explanation

To solve the problem step by step, we will analyze the energy involved when the ball falls and collides with the ground.

### Step 1: Define the variables
Let:
- \( m \) = mass of the ball (we will see that it cancels out)
- \( h = 10 \, \text{m} \) (height from which the ball falls)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( u \) = initial downward velocity of the ball

### Step 2: Calculate the total initial energy
The total initial energy \( E_i \) of the ball when it is at height \( h \) is the sum of its kinetic energy and potential energy:
\[ E_i = \text{K.E.} + \text{P.E.} = \frac{1}{2} m u^2 + mgh \]

### Step 3: Calculate the energy loss during collision
The ball loses 50% of its energy during the collision. Therefore, the energy lost \( E_{loss} \) is:
\[ E_{loss} = \frac{1}{2} E_i = \frac{1}{2} \left( \frac{1}{2} m u^2 + mgh \right) \]

### Step 4: Calculate the remaining energy after collision
The remaining energy \( E_f \) after the collision is:
\[ E_f = E_i - E_{loss} = \frac{1}{2} m u^2 + mgh - \frac{1}{2} \left( \frac{1}{2} m u^2 + mgh \right) \]
This simplifies to:
\[ E_f = \frac{1}{2} m u^2 + mgh - \frac{1}{4} m u^2 - \frac{1}{2} mgh \]
Combining the terms gives:
\[ E_f = \left( \frac{1}{2} - \frac{1}{4} \right) m u^2 + \left( 1 - \frac{1}{2} \right) mgh = \frac{1}{4} m u^2 + \frac{1}{2} mgh \]

### Step 5: Set the remaining energy equal to the potential energy at the height
After the collision, the ball rises back to the same height \( h \), so the remaining energy \( E_f \) is equal to the potential energy at height \( h \):
\[ E_f = mgh \]
Thus, we have:
\[ \frac{1}{4} m u^2 + \frac{1}{2} mgh = mgh \]

### Step 6: Simplify the equation
Dividing through by \( m \) (since \( m \neq 0 \)):
\[ \frac{1}{4} u^2 + \frac{1}{2} gh = gh \]
Rearranging gives:
\[ \frac{1}{4} u^2 = gh - \frac{1}{2} gh \]
\[ \frac{1}{4} u^2 = \frac{1}{2} gh \]

### Step 7: Solve for \( u^2 \)
Multiplying both sides by 4 gives:
\[ u^2 = 2gh \]

### Step 8: Solve for \( u \)
Taking the square root of both sides:
\[ u = \sqrt{2gh} \]

### Step 9: Substitute the values
Substituting \( g = 9.8 \, \text{m/s}^2 \) and \( h = 10 \, \text{m} \):
\[ u = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14 \, \text{m/s} \]

### Final Answer
The initial velocity \( u \) is \( 14 \, \text{m/s} \).

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