A ball dropped from a point P crosses a point Q in t seconds. The time taken by it to travel from Q to R, if PQ = QR.

Options

A. `t`
B. `sqrt(2) t`
C. `2t`
D. `(sqrt(2) - 1) t`

Answer

To solve the problem, we will analyze the motion of the ball dropped from point P to point Q and then from point Q to point R, where PQ = QR.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- A ball is dropped from point P and crosses point Q in time \( t \).
- The distance \( PQ \) is equal to the distance \( QR \).

2. **Using the Equation of Motion**:
- The distance fallen by the ball from P to Q can be calculated using the equation of motion:
\[ PQ = \frac{1}{2} g t^2 \]
- Here, \( g \) is the acceleration due to gravity.

3. **Distance from Q to R**:
- Since \( PQ = QR \), we can denote the distance \( QR \) as \( x \), which is equal to \( PQ \):
\[ QR = \frac{1}{2} g t^2 \]

4. **Total Distance from P to R**:
- The total distance \( PR \) can be expressed as:
\[ PR = PQ + QR = \frac{1}{2} g t^2 + \frac{1}{2} g t^2 = g t^2 \]

5. **Finding the Time from Q to R**:
- Let the time taken to travel from Q to R be \( t_1 \).
- Using the equation of motion again for the distance \( QR \):
\[ QR = \frac{1}{2} g t_1^2 \]
- Since \( QR = \frac{1}{2} g t^2 \), we can set the two equations equal:
\[ \frac{1}{2} g t_1^2 = \frac{1}{2} g t^2 \]
- Simplifying gives:
\[ t_1^2 = t^2 \cdot 2 \]
- Therefore:
\[ t_1 = \sqrt{2} t \]

6. **Finding the Time from Q to R**:
- The total time taken from P to R is:
\[ t + t_1 = t + \sqrt{2} t = t(1 + \sqrt{2}) \]
- The time taken from Q to R is:
\[ t_1 = \sqrt{2} t - t = t(\sqrt{2} - 1) \]

### Final Answer:
The time taken by the ball to travel from Q to R is:
\[ t(\sqrt{2} - 1) \]