A projectile is thrown with velocity v making an angle `theta` with the horizontal. It just crosses the top of two poles,each of height h, after 1 seconds 3 second respectively. The time of flight of the projectile is

Options

A. 2s
B. 6s
C. 8s
D. 4s

Explanation

To solve the problem, we need to analyze the motion of the projectile as it crosses two poles of equal height \( h \) at different times. Let's break down the solution step by step.

### Step 1: Understand the problem
The projectile crosses the first pole after \( t_1 = 1 \) second and the second pole after \( t_2 = 3 \) seconds. We need to find the total time of flight \( T \) of the projectile.

### Step 2: Determine the time interval between the two poles
The time interval between the two poles can be calculated as:
\[ \Delta t = t_2 - t_1 = 3 \, \text{s} - 1 \, \text{s} = 2 \, \text{s} \]

### Step 3: Analyze the motion
The projectile reaches its maximum height at the midpoint of its flight. The time taken to go from the first pole to the maximum height is equal to the time taken to go from the maximum height to the second pole.

### Step 4: Find the time to reach maximum height
Since the time taken to travel from the first pole to the maximum height is equal to the time taken from the maximum height to the second pole, we can denote the time taken to reach maximum height from the first pole as \( t_{sm} \).

Since the total time taken to travel from the first pole to the second pole is \( \Delta t = 2 \, \text{s} \), we have:
\[ t_{sm} = \frac{\Delta t}{2} = \frac{2 \, \text{s}}{2} = 1 \, \text{s} \]

### Step 5: Calculate the total time of flight
The total time of flight \( T \) is the time taken to reach maximum height and then descend back to the ground:
\[ T = t_1 + t_{sm} + t_{sm} = t_1 + 2 \cdot t_{sm} \]
Substituting the known values:
\[ T = 1 \, \text{s} + 2 \cdot 1 \, \text{s} = 1 \, \text{s} + 2 \, \text{s} = 4 \, \text{s} \]

### Conclusion
The total time of flight of the projectile is \( T = 4 \) seconds.

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