From a particular point P above the ground, a ball is thrown vertically upwards. When the ball reaches a particular distance h below point P, its speed is four times its speed at a height h above point P. The maximum height attained by the ball above the point P is

Options

A. `(17)/(15)`h
B. `(17)/(4)` h
C. `(15)/(4)`h
D. `(15)/(17)`h

Answer

To solve the problem step by step, we need to analyze the motion of the ball as it is thrown upwards and then comes downwards.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A ball is thrown vertically upwards from point P.
- When the ball reaches a distance \( h \) below point P, its speed is \( 4V \), where \( V \) is the speed at a height \( h \) above point P.

2. **Setting Up the Equations**:
- Let the initial speed of the ball when thrown from point P be \( U \).
- When the ball reaches height \( h \) above point P, we can use the first equation of motion:
\[ V^2 = U^2 - 2g h \quad \text{(1)} \]
- When the ball reaches a distance \( h \) below point P, its speed is \( 4V \):
\[ (4V)^2 = U^2 + 2g h \quad \text{(2)} \]

3. **Substituting for \( V \)**:
- From equation (1), we can express \( V^2 \):
\[ V^2 = U^2 - 2gh \]
- Substitute \( V^2 \) into equation (2):
\[ 16(U^2 - 2gh) = U^2 + 2gh \]

4. **Expanding and Rearranging**:
- Expanding gives:
\[ 16U^2 - 32gh = U^2 + 2gh \]
- Rearranging terms:
\[ 16U^2 - U^2 = 32gh + 2gh \]
\[ 15U^2 = 34gh \]
- Therefore, we have:
\[ U^2 = \frac{34gh}{15} \quad \text{(3)} \]

5. **Finding Maximum Height**:
- The maximum height \( H \) above point P can be calculated using the formula:
\[ H = \frac{U^2}{2g} \]
- Substituting \( U^2 \) from equation (3):
\[ H = \frac{\frac{34gh}{15}}{2g} = \frac{34h}{30} = \frac{17h}{15} \]

6. **Final Answer**:
- The maximum height attained by the ball above point P is:
\[ H = \frac{17h}{15} \]