Problem 465 - JEE Physics

Question

The potential energy of a projectile at its highest point is (1/2) th the value of its initial kinetic energy. Therefore its angle of projection is (A) \(30^{\circ}\) (B) \(45^{\circ}\) (C) \(60^{\circ}\) (D) \(75^{\circ}\)

Step-by-Step Solution

Answer

The angle of projection is \(\theta = 45^{\circ}\), so the correct option is (B) \(45^{\circ}\).

Step 1: Understand the initial kinetic energy and potential energy at the highest point

At the highest point of the projectile's trajectory, the vertical component of its velocity equals 0. So the kinetic energy is the square of the horizontal component of velocity divided by 2. On the other hand, the potential energy at the highest point is the gained altitude by the projectile multiplied by the gravitational acceleration (g) and its mass (m).

Step 2: Derive the kinetic and potential energy formulas

We know that the speed of the projectile can be written as: \[v = v_0\cos\theta\] The kinetic energy at the highest point will be given by: \[K = \frac{1}{2}mv^2 = \frac{1}{2}m(v_0\cos\theta)^2\] We also know that the height (h) of the highest point can be calculated using the relationship: \[h = \frac{v_0^2\sin^2\theta}{2g}\] The potential energy at the highest point is the product of mass, gravity, and height: \[U = mgh\]

Step 3: Set up the equation and solve for the angle

We are given: \[U = \frac{1}{2}K\] Substitute the equations for kinetic and potential energy: \[\frac{1}{2}m(v_0\cos\theta)^2 = mgh\] Cancel out the mass (m) and rearrange the equation: \[\frac{v_0^2\cos^2\theta}{2} = \frac{v_0^2\sin^2\theta}{2g}\] To eliminate one of the trigonometric functions, we can use the identity: \[\sin^2\theta = 1 - \cos^2\theta\] Substitute this identity into the previous equation: \[\frac{v_0^2\cos^2\theta}{2} = \frac{v_0^2(1-\cos^2\theta)}{2g}\] Simplify the equation: \[\cos^2\theta = 1 - \cos^2\theta\] Solve for \(\cos^2\theta\): \[\cos^2\theta = \frac{1}{2}\] Take the square root to get \(\cos\theta\): \[\cos\theta = \frac{1}{\sqrt{2}}\] Finally, determine the angle of projection, \(\theta\): \[\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right)\] As a result, we obtain the following answer: \[\theta = 45^{\circ}\] So, the correct option is (B) \(45^{\circ}\).

Key Concepts

Kinetic Energy

Kinetic energy is an essential aspect of projectile motion. It refers to the energy that an object possesses due to its motion. For a projectile, its kinetic energy varies as it travels through the air. At the beginning of its journey, when the projectile is launched, its kinetic energy is at its peak.

Kinetic energy can be calculated using the formula:

• \( K = \frac{1}{2}mv^2 \)

where:

• \( m \) is the mass of the object,
• \( v \) is its velocity.

As the projectile ascends, its kinetic energy diminishes, reaching its minimum at the peak of the trajectory due to the vertical component of velocity becoming zero. However, its horizontal kinetic energy remains, as represented by \((v_0 \cos \theta)^2\). This highlights the part of the initial velocity that affects the projectile's range and horizontal motion.

Potential Energy

Potential energy in projectile motion is associated with the object's position in a gravitational field. At the launch of the projectile, its potential energy is lowest. As it gains altitude, the potential energy increases.

It can be expressed using the formula:

• \( U = mgh \)

where:

• \( m \) is the mass of the projectile,
• \( g \) represents the acceleration due to gravity, typically \(9.81 \, m/s^2\),
• \( h \) denotes the height above the launch point.

As the projectile reaches its highest point, it maximizes its potential energy while the kinetic energy in the vertical direction falls to zero. This interplay between kinetic and potential energy is a demonstration of energy conservation in physics. The problem states that the potential energy here is half the value of its initial kinetic energy, which is a crucial clue.

Angle of Projection

The angle of projection determines how a projectile moves through the air. This angle, measured from the horizontal, influences both the range and the height of the projectile.

In our problem, we are tasked with finding the angle at which the potential energy at the projectile’s highest point is half of its initial kinetic energy. By solving equations incorporating trigonometric relations, we found that:

• The cosine of the angle squared is half.

The relation \( \cos^2 \theta = \frac{1}{2} \) yields the angle \( \theta \) as \( 45^\circ \).

This result reflects the balance between horizontal and vertical components of velocity, where each component contributes equally to the projectile's initial velocity, thus maximizing the range for given initial velocity conditions.

Trigonometric Identities

Understanding trigonometric identities is crucial to dissecting projectile motion and solving related problems. The identity used in this problem is:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This identity allows us to express one trigonometric function in terms of another and is vital in solving for unknown angles in projectile motion.

In the equation setup:

• By substituting \( \sin^2 \theta = 1 - \cos^2 \theta \), we formed an equation in terms of \( \cos \theta \) solely, simplifying the calculation substantially.

Recognizing and applying these identities is invaluable, not just for learning projectile motion, but for tackling many branches of physics and mathematics that require an understanding of angles and ratios.