It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point at a distance r from the point of projection. The product of times taken to reach this point in the two possible ways is then proportional to

Options

A. r
B. `r^(2)`
C. `1//r`
D. `1//r^(2)`

Explanation

To solve the problem, we need to analyze the projectile motion of a particle projected at two different angles that allows it to pass through the same point at a distance \( r \) from the point of projection. We will derive the relationship between the product of the times taken to reach that point and the distance \( r \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to project a particle with a given initial velocity \( u \) such that it passes through a point at a distance \( r \) from the projection point. The particle can be projected at two different angles, \( \theta \) and \( 90^\circ - \theta \).

2. **Range of Projectile Motion**:
The range \( R \) of a projectile launched at an angle \( \theta \) is given by the formula:
\[ R = \frac{u^2 \sin(2\theta)}{g} \]
Since both projections must reach the same point \( P \) at distance \( r \), we can equate the range to \( r \):
\[ r = \frac{u^2 \sin(2\theta)}{g} \]

3. **Time of Flight**:
The time of flight \( t \) for a projectile launched at an angle \( \theta \) is given by:
\[ t = \frac{2u \sin(\theta)}{g} \]
For the two angles \( \theta \) and \( 90^\circ - \theta \), we have:
- For angle \( \theta \):
\[ t_1 = \frac{2u \sin(\theta)}{g} \]
- For angle \( 90^\circ - \theta \):
\[ t_2 = \frac{2u \sin(90^\circ - \theta)}{g} = \frac{2u \cos(\theta)}{g} \]

4. **Product of Times**:
Now, we calculate the product of the times taken to reach point \( P \):
\[ t_1 \cdot t_2 = \left(\frac{2u \sin(\theta)}{g}\right) \cdot \left(\frac{2u \cos(\theta)}{g}\right) \]
Simplifying this gives:
\[ t_1 \cdot t_2 = \frac{4u^2 \sin(\theta) \cos(\theta)}{g^2} \]

5. **Using the Identity**:
We can use the trigonometric identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \):
\[ t_1 \cdot t_2 = \frac{2u^2 \sin(2\theta)}{g^2} \]

6. **Relating to Distance \( r \)**:
From the earlier step, we know that:
\[ r = \frac{u^2 \sin(2\theta)}{g} \]
Therefore, we can express \( \sin(2\theta) \) in terms of \( r \):
\[ \sin(2\theta) = \frac{rg}{u^2} \]
Substituting this back into the product of times gives:
\[ t_1 \cdot t_2 = \frac{2u^2}{g^2} \cdot \frac{rg}{u^2} = \frac{2r}{g} \]

7. **Conclusion**:
Thus, we find that the product of the times taken to reach the point \( P \) is directly proportional to \( r \):
\[ t_1 \cdot t_2 \propto r \]

### Final Answer:
The product of times taken to reach the point is proportional to \( r \).