A ball is thrown vertically upwards with a velocity of 10 `ms^(-1)`. It returns to the ground with a velocity of 9 `ms^(-1)`. If g=9.8 `ms^(-2)` , then the maximum height attained by the ball is nearly (assume air resistance to be uniform)

Options

A. 5.1 m
B. 4.1 m
C. 4.61 m
D. 5 m

Explanation

To find the maximum height attained by the ball thrown vertically upwards, we can use the equations of motion along with the concepts of forces acting on the ball during its ascent and descent.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial velocity (u) = 10 m/s (upwards)
- Final velocity (v) when returning to the ground = 9 m/s (downwards)
- Acceleration due to gravity (g) = 9.8 m/s²

2. **Understand the Forces Acting on the Ball**:
- When the ball is thrown upwards, it experiences two forces: gravitational force (mg) acting downwards and air resistance (r) acting downwards as well. The net acceleration (A1) while going up is:
\[ A_1 = g + \frac{r}{m} \]
- When the ball is coming down, it experiences gravitational force (mg) acting downwards and air resistance (r) acting upwards. The net acceleration (A2) while coming down is:
\[ A_2 = g - \frac{r}{m} \]

3. **Using the First Equation of Motion for Ascent**:
- When the ball reaches its maximum height, its final velocity (v) is 0 m/s. Using the equation:
\[ v^2 = u^2 - 2A_1H \]
- Substituting the values:
\[ 0 = (10)^2 - 2(g + \frac{r}{m})H \]
- This simplifies to:
\[ 100 = 2(g + \frac{r}{m})H \quad \text{(Equation 1)} \]

4. **Using the First Equation of Motion for Descent**:
- When the ball returns to the ground, we can use:
\[ v^2 = u^2 + 2A_2H \]
- Here, the initial velocity (u) at the maximum height is 0 m/s, and the final velocity (v) is 9 m/s:
\[ (9)^2 = 0 + 2(g - \frac{r}{m})H \]
- This simplifies to:
\[ 81 = 2(g - \frac{r}{m})H \quad \text{(Equation 2)} \]

5. **Adding Both Equations**:
- Adding Equation 1 and Equation 2:
\[ 100 + 81 = 2(g + \frac{r}{m})H + 2(g - \frac{r}{m})H \]
- This simplifies to:
\[ 181 = 4gH \]
- Rearranging gives:
\[ H = \frac{181}{4g} \]

6. **Substituting the Value of g**:
- Now substituting g = 9.8 m/s²:
\[ H = \frac{181}{4 \times 9.8} \approx \frac{181}{39.2} \approx 4.61 \text{ meters} \]

### Final Answer:
The maximum height attained by the ball is approximately **4.61 meters**.