A ball is thrown vertically with some velocity . A constant air resistance acts. If the time of ascent is `t_(1)` and that of descent is `t_(2)` , then

Options

A. `t_(1) lt t_(2)`
B. `t_(1) gt t_(2)`
C. `t_(1) = t_(2)`
D. None

Explanation

To solve the problem, we need to analyze the motion of a ball thrown vertically upwards with an initial velocity \( u \) while considering the effects of air resistance. The time taken for the ball to ascend is \( t_1 \) and the time taken to descend is \( t_2 \). We want to find the relationship between \( t_1 \) and \( t_2 \).

### Step-by-Step Solution:

1. **Understanding 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. Thus, the net force acting on the ball during ascent is \( F = mg + r \).
- During descent, the gravitational force still acts downwards, but the air resistance acts upwards. Therefore, the net force during descent is \( F = mg - r \).

2. **Acceleration During Ascent and Descent**:
- For ascent, the acceleration \( a_1 \) can be expressed as:
\[ a_1 = \frac{F}{m} = g + \frac{r}{m} \]
- For descent, the acceleration \( a_2 \) is given by:
\[ a_2 = \frac{F}{m} = g - \frac{r}{m} \]

3. **Using Equations of Motion**:
- For ascent, using the equation \( v = u - a_1 t_1 \) and knowing that the final velocity \( v = 0 \) at the highest point:
\[ 0 = u - (g + \frac{r}{m}) t_1 \]
Rearranging gives:
\[ u = (g + \frac{r}{m}) t_1 \quad \text{(1)} \]
- For descent, using the equation \( h = u t_2 + \frac{1}{2} a_2 t_2^2 \) and knowing that the initial velocity for descent is \( 0 \):
\[ h = \frac{1}{2} (g - \frac{r}{m}) t_2^2 \quad \text{(2)} \]

4. **Equating Heights**:
- The height \( h \) reached during ascent and descent is the same. Thus, we can equate equations (1) and (2):
\[ (g + \frac{r}{m}) t_1^2 = \frac{1}{2} (g - \frac{r}{m}) t_2^2 \]

5. **Dividing the Equations**:
- Dividing both sides by \( t_2^2 \) and rearranging gives:
\[ \frac{t_1^2}{t_2^2} = \frac{(g - \frac{r}{m})}{2(g + \frac{r}{m})} \]

6. **Taking Square Roots**:
- Taking the square root of both sides, we find:
\[ \frac{t_1}{t_2} = \sqrt{\frac{(g - \frac{r}{m})}{2(g + \frac{r}{m})}} \]

7. **Analyzing the Result**:
- Since \( g - \frac{r}{m} < g + \frac{r}{m} \), it follows that:
\[ \frac{t_1}{t_2} < 1 \implies t_1 < t_2 \]

### Conclusion:
The time of ascent \( t_1 \) is less than the time of descent \( t_2 \). Therefore, the final relationship is:
\[ t_1 < t_2 \]