Problem 54 - Master Resource Book in JEE Main Physics

Question

A particle is thrown in the upward direction making an angle of \(60^{\circ}\) with the horizontal direction with a velocity of \(147 \mathrm{~ms}^{-1}\). Then, the time after which its inclination with the horizontal is \(45^{\circ}\), is IUP SEE 2006] (a) \(15 \mathrm{~s}\) (b) \(10.98 \mathrm{~s}\) (c) \(5.49 \mathrm{~s}\) (d) \(2.745 \mathrm{~s}\)

Step-by-Step Solution

Answer

The time is approximately 5.49 seconds, which corresponds to option (c).

Step 1: Break Down Initial Velocity

The particle is thrown with an initial velocity of \(147 \, \mathrm{ms}^{-1}\) at an angle of \(60^\circ\). This means we split the velocity into horizontal and vertical components. Use the formulas: \[ V_{x} = V_0 \cos \theta \] and \[ V_{y} = V_0 \sin \theta \]where \( V_0 = 147 \, \mathrm{ms}^{-1} \) and \( \theta = 60^\circ \).Calculate the horizontal component, \[ V_{x} = 147 \cos 60^\circ = 147 \times \frac{1}{2} = 73.5 \, \mathrm{ms}^{-1} \].Calculate the vertical component, \[ V_{y} = 147 \sin 60^\circ = 147 \times \frac{\sqrt{3}}{2} = 127.2792 \, \mathrm{ms}^{-1} \].

Step 2: Understand the Inclination Change

You want to find the time when the angle of inclination changes from \(60^\circ\) to \(45^\circ\). The tangent of the angle \(\theta\) gives the relationship: \[ \tan \theta = \frac{V_{y}}{V_{x}} \]. Initially, \(\tan 60^\circ = \frac{127.2792}{73.5}\).

Step 3: Set Up for Angle 45° Calculation

At inclination \(45^\circ\), \(\tan 45^\circ = 1\). Set up this equation using the new vertical velocity, adjusted for the gravitational pull:\[ V_{y}' = V_{y} - gt \], where \(g = 9.8 \, \mathrm{ms}^{-2}\) and \(t\) is the time in seconds.

Step 4: Solve for Time

Using the equation from the previous step:\[ \tan 45^\circ = \frac{V_{y} - gt}{V_{x}} = 1 \].So, \[ 73.5 = 127.2792 - 9.8t \].Rearrange and solve for \(t\):\[ 9.8t = 127.2792 - 73.5 \]\[ 9.8t = 53.7792 \]\[ t = \frac{53.7792}{9.8} \approx 5.49 \mathrm{~s} \].

Step 5: Verify the Answer

The calculated time is approximately \(5.49 \mathrm{~s}\). This matches with one of the given choices, confirming that the time after which the particle's inclination with the horizontal is \(45^\circ\) is \(5.49 \mathrm{~s}\).

Key Concepts

Angular Inclination

When dealing with projectile motion, understanding **angular inclination** is crucial. Angular inclination refers to the angle at which a projectile is launched concerning a reference direction, typically the horizontal. In our exercise, the particle is launched at an angle of \(60^{\circ}\) with the horizontal.

Angular inclination is essential because it affects the projectile's motion in the air. The initial launch angle determines the trajectory, effectively splitting the velocity into two components—horizontal and vertical. These components help us track the path of the projectile as it moves through the air.

In the problem under consideration, we seek the time it takes for the launch angle to change to \(45^{\circ}\). This requires understanding how the angle evolves over time due to gravitational effects acting on the vertical velocity component.

Velocity Components

The motion of any projectile can be described by its **velocity components**—horizontal and vertical. At launch, the initial velocity is divided into these two components using trigonometric functions.
- **Horizontal component** \(V_x\) remains constant in the absence of air resistance: \[ V_{x} = V_0 \cos \theta \] where \(V_0 = 147 \, \mathrm{ms}^{-1}\) and \(\theta = 60^{\circ}\).- **Vertical component** \(V_y\) is affected by gravity: \[ V_{y} = V_0 \sin \theta \]After breaking down the velocity, we find the horizontal component to be \(73.5 \, \mathrm{ms}^{-1}\), and the vertical component is \(127.2792 \, \mathrm{ms}^{-1}\).
Understanding these components allows us to calculate the changes in the vertical velocity over time, especially how it decreases due to gravitational force. This will later be critical in determining when the angle of the projectile with respect to the horizontal changes to a specified value.

Time of Flight

The **time of flight** for a projectile is the duration it remains in the air. It consists of the upward and downward journey of the projectile. In our specific problem, we're interested in the time it takes for the projectile's inclination to change from \(60^{\circ}\) to \(45^{\circ}\).
To find this, we use the relationship between the components of velocity and the angle of inclination, given by:- \[ \tan \theta = \frac{V_{y}}{V_{x}} \] This equation helps us monitor the vertical change in velocity due to gravity over time:\[ V_{y}' = V_{y} - gt \]Where:- \(g\) is the acceleration due to gravity \(9.8 \, \mathrm{ms}^{-2}\) - \(t\) is the time Set \(\tan 45^{\circ} = 1\), and solve for time:\[ \frac{V_{y} - gt}{V_{x}} = 1 \]This provides the point when the inclination reaches \(45^{\circ}\). Solving these equations reveals that the time needed is approximately \(5.49\) seconds. This duration represents the time when gravity affects the trajectory sufficiently to alter the inclination angle to \(45^{\circ}\).