Two particles A and B are projected simultaneously in horizontal direction with speed `50 m//s` and `100m//s` respectively . If `t_(A) and t_(B)` are the time taken by the projectiles to hit the ground by the particles A and B respectively, then

Options

A. `t_(A)=t_(B)`
B. `t_(a) gt t_(B)`
C. `t_(A) lt t_(B)`
D. Information is insufficient to compare `t_(A) and t_(B)`.

Explanation

To solve the problem, we need to analyze the motion of both particles A and B. Since both particles are projected horizontally, the time taken for each particle to hit the ground depends solely on the vertical motion, which is influenced by gravity.

### Step-by-Step Solution:

1. **Understanding the Vertical Motion**:
- Both particles A and B are projected horizontally. The only force acting on them in the vertical direction is gravity (g = 9.81 m/s²).
- The initial vertical velocity (u_y) for both particles is 0 m/s since they are projected horizontally.

2. **Using the Equation of Motion**:
- The time taken to hit the ground can be derived from the second equation of motion:
\[ s = u_y t + \frac{1}{2} g t^2 \]
- Here, \(s\) is the vertical distance fallen (which is the same for both particles when they hit the ground), \(u_y\) is the initial vertical velocity (0 m/s), \(g\) is the acceleration due to gravity, and \(t\) is the time taken to hit the ground.

3. **Simplifying the Equation**:
- Since \(u_y = 0\), the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]
- Rearranging gives:
\[ t^2 = \frac{2s}{g} \]
- Thus, the time \(t\) can be expressed as:
\[ t = \sqrt{\frac{2s}{g}} \]

4. **Conclusion About Time Taken**:
- Since both particles A and B fall the same vertical distance \(s\) and experience the same acceleration due to gravity \(g\), the time taken \(t_A\) for particle A and \(t_B\) for particle B to hit the ground will be the same:
\[ t_A = t_B \]

### Final Answer:
The time taken by both particles A and B to hit the ground is the same:
\[ t_A = t_B \]