A ball is throw vertically upward. It has a speed of `10 m//s` when it has reached on half of its maximum height. How high does the ball rise ? (Taking `g = 10 m//s^2`).

Options

A. 5m
B. 15m
C. 10m
D. 20m

Explanation

To solve the problem of how high the ball rises when thrown vertically upward with an initial speed of `10 m/s` at half of its maximum height, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The ball is thrown vertically upward.
- It reaches a speed of `10 m/s` at half of its maximum height (`H/2`).
- We need to find the maximum height `H`.

2. **Using the Third Equation of Motion**:
- At the maximum height, the final velocity (`v`) is `0 m/s`.
- The initial velocity (`u`) at half the maximum height is `10 m/s`.
- The acceleration due to gravity (`g`) is `10 m/s²` (acting downwards).

3. **Applying the Equation**:
- We can use the third equation of motion:
\[ v^2 = u^2 - 2gH \]
- For the maximum height, we can set this up as:
\[ 0 = u^2 - 2gH \]
- Rearranging gives:
\[ u^2 = 2gH \quad \text{(Equation 1)} \]

4. **Finding the Velocity at Half the Maximum Height**:
- Now, we consider the motion when the ball is at half the maximum height (`H/2`).
- Here, the initial velocity (`u`) is `10 m/s` and the final velocity (`v`) is also `10 m/s` (the speed when it reaches `H/2`).
- We use the same equation:
\[ v^2 = u^2 - 2g\left(\frac{H}{2}\right) \]
- Plugging in the values gives:
\[ 10^2 = u^2 - gH \]
- This simplifies to:
\[ 100 = u^2 - gH \quad \text{(Equation 2)} \]

5. **Substituting Equation 1 into Equation 2**:
- From Equation 1, we know:
\[ u^2 = 2gH \]
- Substitute this into Equation 2:
\[ 100 = 2gH - gH \]
- This simplifies to:
\[ 100 = gH \]

6. **Solving for Maximum Height (H)**:
- Now, substituting the value of `g`:
\[ 100 = 10H \]
- Dividing both sides by `10` gives:
\[ H = 10 \text{ meters} \]

### Final Answer:
The maximum height to which the ball rises is **10 meters**.