Problem 52 - Master Resource Book in JEE Main Physics

Question

A particle is projected with velocity \(2 \sqrt{g h}\) so that it just clears two walls of equal height \(h\), which are at a distance of \(2 h\) from each other. What is the time interval of passing between the two walls? (a) \(\frac{2 h}{g}\) (b) \(\sqrt{\frac{g h}{g}}\) (c) \(\sqrt{\frac{h}{g}}\) (d) \(2 \sqrt{\frac{h}{g}}\)

Step-by-Step Solution

Step 1: Initial Velocity of the Particle

The particle is projected with an initial velocity \( v_0 = 2 \sqrt{g h} \). This velocity will affect how the particle moves over time.

Step 2: Horizontal and Vertical Components of the Velocity

Since we need to analyze the motion in both horizontal and vertical directions, we split the velocity into components. Assume the angle of projection is \( \theta \). The horizontal component is \( v_{0x} = v_0 \cos(\theta) \) and the vertical component is \( v_{0y} = v_0 \sin(\theta) \).

Key Concepts

Time of Flight

The time of flight in projectile motion is the duration for which the particle remains in the air. It's important as it determines how long an object stays airborne before returning to the surface. In the exercise, we are interested in the time interval of the particle passing between two walls.
To find this, we consider the motion's symmetry and balance in the horizontal direction, as the distance between the walls and the velocity's horizontal component help determine the time spent in the air between them. If the time to reach the first wall is known, understanding the time spent covering the distance between both walls becomes straightforward.
This time is calculated using the formula for horizontal motion: \[t = \frac{d}{v_{0x}}\]Where:

• \(d\) is the horizontal distance between the walls.
• \(v_{0x}\) is the horizontal component of the velocity.

Substitute the distance as \(2h\) and solve for \(t\) to find the time interval of passing between the walls.

Components of Velocity

Understanding the components of velocity is crucial in analyzing projectile motion. When a particle is projected, its initial velocity is split into two components: horizontal and vertical. These components influence how the particle travels through space.
The horizontal component, \(v_{0x}\), is responsible for the particle's movement along the horizontal axis. It's calculated by multiplying the initial velocity by the cosine of the angle of projection: \[v_{0x} = v_0 \cos(\theta)\]This component remains constant, as there is no acceleration in the horizontal direction in ideal conditions.
The vertical component, \(v_{0y}\), affects the particle's motion along the vertical axis and is given by:\[v_{0y} = v_0 \sin(\theta)\]This component changes throughout the particle's flight due to gravitational acceleration. The rising and falling paths result from this acceleration, causing the particle to ascend to a peak height and then descend.

Angle of Projection

The angle of projection, often denoted as \(\theta\), is the angle at which an object is launched into the air. It's a vital factor in determining the trajectory and range of a projectile.
This angle influences both the time of flight and the maximum height reached by the projectile. By altering the angle, you can change the balance between the horizontal and vertical components of the initial velocity. A standard angle for achieving maximum range on a level surface is \(45^\circ\), as it perfectly balances these components.
In our problem, knowing the angle is essential for correctly calculating the time to clear the walls. The relationship between the angle, initial velocity, and the distance it needs to travel is quite direct:

• The sharper the angle, the higher the initial vertical component, leading to a higher but shorter trajectory.
• A shallower angle results in a smaller vertical component, giving longer but lower trajectories.

By understanding the influence of the projection angle, students can better predict and calculate the particle's path.