A glass ball is dropped from height 10 m. If there is 15% loss of energy due to impact, then after one impact, the ball will go upto.

Options

A. 2m
B. 4m
C. 6m
D. None of these

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the initial velocity just before impact
When the glass ball is dropped from a height of 10 m, we can use the equation of motion to find the velocity just before impact. The equation is:

\[ v^2 = u^2 + 2gh \]

Where:
- \( v \) = final velocity just before impact
- \( u \) = initial velocity (which is 0 since the ball is dropped)
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))
- \( h \) = height (10 m)

Substituting the values:

\[ v^2 = 0 + 2 \times 10 \times 10 \]
\[ v^2 = 200 \]
\[ v = \sqrt{200} = 10\sqrt{2} \, \text{m/s} \]

### Step 2: Calculate the initial kinetic energy before impact
The kinetic energy (KE) just before impact can be calculated using the formula:

\[ KE = \frac{1}{2} mv^2 \]

Substituting \( v^2 = 200 \):

\[ KE = \frac{1}{2} m \times 200 = 100m \, \text{J} \]

### Step 3: Calculate the energy loss due to impact
The problem states that there is a 15% loss of energy due to impact. Therefore, the energy lost is:

\[ \text{Energy lost} = 0.15 \times KE = 0.15 \times 100m = 15m \, \text{J} \]

### Step 4: Calculate the final kinetic energy after impact
The final kinetic energy after the impact will be:

\[ KE_{\text{final}} = KE - \text{Energy lost} \]
\[ KE_{\text{final}} = 100m - 15m = 85m \, \text{J} \]

### Step 5: Calculate the final velocity after impact
Using the final kinetic energy to find the final velocity \( v' \):

\[ KE_{\text{final}} = \frac{1}{2} mv'^2 \]
\[ 85m = \frac{1}{2} mv'^2 \]

Cancelling \( m \) from both sides:

\[ 85 = \frac{1}{2} v'^2 \]
\[ v'^2 = 170 \]
\[ v' = \sqrt{170} \, \text{m/s} \]

### Step 6: Calculate the maximum height reached after the impact
The maximum height \( h' \) reached after the impact can be calculated using the formula:

\[ h' = \frac{v'^2}{2g} \]

Substituting \( v'^2 = 170 \) and \( g = 10 \):

\[ h' = \frac{170}{2 \times 10} = \frac{170}{20} = 8.5 \, \text{m} \]

### Conclusion
After one impact, the ball will go up to a maximum height of **8.5 meters**.

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