A particle is projected with velocity 20 m/s , so that it just clears two walls of equal height 10 m, which are at a distance 20 m from each other. The time of passing between the wall is

To determine the time a particle takes to pass between two walls, we first need to analyze the motion of the particle under the effect of gravity. The problem states that the particle is projected with an initial velocity of 20 m/s and must clear two walls that are 10 m high and spaced 20 m apart. We’ll break down the solution into manageable steps.

Understanding the Motion

The particle's motion can be divided into horizontal and vertical components. The initial velocity can be separated into these components based on the angle of projection. However, since the angle is not specified, we'll assume the projection is horizontal for simplicity, as this gives us a straightforward path to analyze.

Vertical Motion Analysis

The vertical distance the particle must clear is 10 m. The vertical motion is influenced by gravity, which we denote as \( g = 9.81 \, \text{m/s}^2 \). To find the time it takes to reach this height, we can use the second equation of motion:

The equation is given by:

• \( h = v_{y} t - \frac{1}{2} g t^2 \)

In our case, we want to find the time \( t \) when the particle reaches a vertical height \( h \) of 10 m, and we can assume that the vertical component of the initial velocity \( v_{y} \) is 0 (since we are considering a horizontal projection). The equation simplifies to:

• \( 10 = 0 - \frac{1}{2} \cdot 9.81 \cdot t^2 \)

Rearranging gives:

• \( \frac{1}{2} \cdot 9.81 \cdot t^2 = 10 \)

Solving for \( t^2 \) yields:

• \( t^2 = \frac{20}{9.81} \)

Calculating this gives:

• \( t^2 \approx 2.04 \)

Thus:

• \( t \approx \sqrt{2.04} \approx 1.43 \, \text{s} \)

Horizontal Motion Analysis

Next, we need to analyze the horizontal motion. The horizontal distance between the walls is 20 m, and the particle's horizontal velocity is 20 m/s (assuming the projection is horizontal). The time \( t_h \) to travel this distance can be calculated using:

• \( d = v_{x} \cdot t_h \)

Substituting in the values gives:

• \( 20 = 20 \cdot t_h \)

This simplifies to:

• \( t_h = 1 \, \text{s} \)

Final Consideration

Since the time to clear the vertical height of the wall (1.43 s) is greater than the time to travel horizontally between the walls (1 s), the particle will indeed clear the walls as it travels horizontally. Therefore, the time taken to pass between the walls is 1 second, as this is the time required to cover the horizontal distance.

In summary, while the particle takes approximately 1.43 seconds to reach the height of 10 m, it only needs 1 second to move horizontally between the walls. Thus, the time of passing between the walls is 1 second.