Problem 30 - Master Resource Book in JEE Main Physics

Question

The kinetic energy of a project at the height point is half of the initial kinetic energy. What is the angle of projection with the horizontal? [a) \(3 \overline{0^{\circ}}\) (b) \(45^{*}\) (c) \(6 \overline{0^{n}}\) (d) \(90^{*}\)

Step-by-Step Solution

Answer

The angle of projection is \(45^{\circ}\).

Step 1: Understanding the problem

The problem states that the kinetic energy at the highest point is half of the initial kinetic energy. We need to find the angle of projection with the horizontal such that this condition is satisfied.

Step 2: Relate kinetic energy to velocities

The kinetic energy is given by \( KE = \frac{1}{2}mv^2 \). At the highest point, the vertical component of velocity is zero, so only the horizontal component contributes to kinetic energy.

Step 3: Identify the relation between initial and highest point velocities

Initially, the velocity has both horizontal and vertical components: \(v_x = v_0 \cos \theta\) and \(v_y = v_0 \sin \theta\). At the top, the kinetic energy is due to \(v_x\) which is \(v_0 \cos \theta\).

Step 4: Set up the kinetic energy ratio equation

The kinetic energy at the top is \(\frac{1}{2} m (v_0 \cos \theta)^2\) and initially it is \(\frac{1}{2} m v_0^2\). The kinetic energy at the top is half of the initial kinetic energy: \((1/2) m (v_0 \cos \theta)^2 = \frac{1}{2}(1/2) m v_0^2\).

Step 5: Solve for \( \cos \theta \)

Simplify the equation: \((v_0 \cos \theta)^2 = \frac{1}{2} v_0^2\), which simplifies to \(\cos^2 \theta = \frac{1}{2}\). This implies \(\cos \theta = \frac{1}{\sqrt{2}}\).

Step 6: Identify the angle of projection

\(\cos \theta = \frac{1}{\sqrt{2}}\) corresponds to \(\theta = 45^{\circ}\). Therefore, the angle of projection is \(45^{\circ}\).

Key Concepts

Kinetic Energy

Kinetic energy is a fundamental concept in physics, especially when discussing projectile motion. It represents the energy that an object possesses due to its motion. In mathematical terms, kinetic energy is described by the formula: \( KE = \frac{1}{2}mv^2 \). Here, \( m \) is the mass of the object, and \( v \) is its velocity. This formula shows that kinetic energy depends directly on the square of the velocity. This quadratic relationship means that even small increases in velocity will significantly increase kinetic energy.
When dealing with projectile motion, it's crucial to understand how kinetic energy varies throughout the motion. For instance, at the highest point in a projectile's path, the kinetic energy is at a minimum because the vertical component of velocity is zero, leaving only the horizontal component. As seen in our specific problem, if a projectile's kinetic energy at the peak is half of its initial kinetic energy, this inference allows determining the angle of projection if other parameters are known.

Angle of Projection

The angle of projection is vital in determining a projectile's trajectory. This angle is the initial angle formed between the velocity vector and the horizontal axis. Understanding this angle is crucial since it influences the path, range, and maximum height reached by the projectile.
In many physics problems, finding the correct angle of projection can help solve for unknowns, such as maximum height or range. In the exercise example, knowing that the kinetic energy at the highest point is half of the initial kinetic energy helps us deduce the angle to be \(45^{\circ}\). Here's how:

• The kinetic energy equation at the top is derived using the horizontal velocity component \(v_0 \cos \theta\).
• The statement \( \cos \theta = \frac{1}{\sqrt{2}} \) points us directly to the \(45^{\circ}\) solution because this value of cosine precisely corresponds to \(45^{\circ}\).

This particular angle not only satisfies the kinetic condition given but is also a common angle resulting in a perfect balance between the horizontal and vertical velocity components in projectile motion.

Velocity Components

An integral part of understanding projectile motion is breaking down the initial velocity into its horizontal and vertical components. These components are derived from the initial speed and the angle of projection.

• **Horizontal Component**: The horizontal velocity component is given by \( v_x = v_0 \cos \theta \), which remains constant throughout the projectile's flight in the absence of air resistance, as it is not influenced by gravity.
• **Vertical Component**: Conversely, the vertical component is \( v_y = v_0 \sin \theta \). This component is affected by gravity, which acts to slow down the projectile until it reaches its peak and then speeds it up as it falls back to the ground.

These components are crucial for determining the projectile's kinetic energy at various points in its path.
In the exercise mentioned, at the highest point, the vertical component \( v_y \) becomes zero, leaving only the horizontal component \( v_x \) to contribute to kinetic energy. This simplification helps us solve for the angle \( \theta \), showing how deep the interplay between velocity components and projection angles is in solving physics problems.