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 the two poles of equal height \( h \) at different times. The key points are:

1. The projectile crosses the first pole at \( t_1 = 1 \) second.
2. The projectile crosses the second pole at \( t_2 = 3 \) seconds.
3. Both poles are of the same height \( h \).

### Step-by-Step Solution:

**Step 1: Understanding the vertical motion of the projectile.**

The vertical displacement of the projectile can be described by the equation of motion:
\[ y = v_{y0} t - \frac{1}{2} g t^2 \]
where:
- \( y \) is the vertical displacement (height of the poles, \( h \)),
- \( v_{y0} = v \sin(\theta) \) is the initial vertical component of the velocity,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( t \) is the time.

**Step 2: Setting up the equations for the two poles.**

For the first pole at \( t_1 = 1 \) second:
\[ h = v \sin(\theta) \cdot 1 - \frac{1}{2} g \cdot (1)^2 \]
This simplifies to:
\[ h = v \sin(\theta) - \frac{1}{2} g \]

For the second pole at \( t_2 = 3 \) seconds:
\[ h = v \sin(\theta) \cdot 3 - \frac{1}{2} g \cdot (3)^2 \]
This simplifies to:
\[ h = 3v \sin(\theta) - \frac{9}{2} g \]

**Step 3: Setting the two equations equal to each other.**

Since both expressions equal \( h \), we can set them equal:
\[ v \sin(\theta) - \frac{1}{2} g = 3v \sin(\theta) - \frac{9}{2} g \]

**Step 4: Rearranging the equation.**

Rearranging gives:
\[ - \frac{1}{2} g + \frac{9}{2} g = 3v \sin(\theta) - v \sin(\theta) \]
\[ 4g = 2v \sin(\theta) \]
\[ 2g = v \sin(\theta) \]

**Step 5: Finding the time of flight.**

The time of flight \( T \) for a projectile launched at an angle \( \theta \) can be calculated using the formula:
\[ T = \frac{2v \sin(\theta)}{g} \]

Substituting \( v \sin(\theta) = 2g \) into the time of flight formula:
\[ T = \frac{2(2g)}{g} = 4 \text{ seconds} \]

### Final Answer:
The time of flight of the projectile is \( T = 4 \) seconds.

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