Question

When a particle is projected from the surface of earth, it mechanical energy and angular momentum about center of earth at all time is constant (i) A particle of mass \(\mathrm{m}\) is projected from the surface of earth with velocity \(\mathrm{V}_{0}\) at angle \(\theta\) with horizontal suppose \(\mathrm{h}\) be the maximum height of particle from surface of earth and \(\mathrm{v}\) its speed at that point them \(\mathrm{V}\) is (A) \(\mathrm{V}_{0} \cos \theta\) \((\mathrm{B})>\mathrm{V}_{0} \cos \theta\) (C) \(<\mathrm{V}_{0} \cos \theta\) (D) zero (ii) Maximum height h of the particle is \((\mathrm{A})=\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]\) (B) \(>\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]\) \((\mathrm{C})<\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]\) (D) can be greater than or less than \(\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]\)

Short answer

The short answer is as follows: For the particle's velocity at maximum height, the correct option is (A): \(V = V_0 \cos \theta\). For the maximum height h, the correct option is (C): \(h < \frac{V_0^2 \sin^2 \theta}{2g}\).

Step-by-step solution

Identify the Components of Initial Velocity

First, we need to split the initial velocity, V_0, into its horizontal (V_{0x}) and vertical (V_{0y}) components: \(V_{0x} = V_0 \cos \theta\) \(V_{0y} = V_0 \sin \theta\)

Determine the Expression for Mechanical Energy and Angular Momentum

Due to the conservation of mechanical energy (TME = kinetic energy (K) + potential energy (U)), angular momentum (L) equation for the particle should look like this: \(TME = K + U = \frac{1}{2} m v^2 + mgh\) where \(L = mvr\).

Calculate Velocity at Maximum Height

At maximum height (h), the vertical component of the particle's velocity is V_y = 0. The horizontal component, V_x, remains constant because no external forces are acting on it. Therefore, particle's speed at the maximum height (V) is equal to the horizontal component of the initial velocity: \(V = V_{0x}\) According to the given choices: (A) \(V = V_0 \cos \theta\) (B) \(V > V_0 \cos \theta\) (C) \(V < V_0 \cos \theta\) (D) 0 V = V_0*cos(θ) is the correct answer for the particle's velocity at maximum height. (Option A)

Calculate the Maximum Height (h)

At the maximum height h, the potential energy U (due to height) is entirely converted into kinetic energy K (due to horizontal motion). From the conservation of mechanical energy (TME) equation: \(\frac{1}{2} m V^2 = mgh\) Using the value of V calculated from Step 3: \(\frac{1}{2} m V_{0x}^2 = mgh\) Now, solve for h: \(h = \frac{V_{0x}^2}{2g}\) \(h = \frac{(V_0 \cos \theta)^2}{2g}\) Comparing this to the given choices: (A) \(h = \frac{V_0^2 \sin^2 \theta}{2g}\) (B) \(h > \frac{V_0^2 \sin^2 \theta}{2g}\) (C) \(h < \frac{V_0^2 \sin^2 \theta}{2g}\) (D) \(h\) can be greater than or less than \(\frac{V_0^2 \sin^2 \theta}{2g}\) Since \(h = \frac{(V_0 \cos \theta)^2}{2g}\), we can say that \(h < \frac{V_0^2 \sin^2 \theta}{2g}\). (Option C)

Key concepts

Mechanical Energy Conservation

Mechanical energy conservation is a foundational concept in physics. It states that in a closed system, without external forces, the total mechanical energy remains constant. This total mechanical energy comprises kinetic energy and potential energy. In projectile motion, this principle helps us understand how the energy transitions between these forms.

For instance, when a projectile is launched, it has maximum kinetic energy at the start because of its velocity. As it rises, some of this kinetic energy converts into potential energy, defined by its height. At the peak height, the energy is mostly potential. As it falls back down, this potential energy turns back into kinetic energy.

The equation for mechanical energy conservation in this context is given as:

• Total Mechanical Energy (TME) = Kinetic Energy (K) + Potential Energy (U)
• Kinetic Energy, K = \( \frac{1}{2} m v^2 \)
• Potential Energy, U = \( mgh \)

This knowledge is essential for solving problems related to projectile motion, particularly when calculating velocities and heights at various points in the projectile's trajectory.

Angular Momentum

Angular momentum is another key concept that deals with the rotational motion of objects. For a particle moving with velocity in a circle, its angular momentum is constant about a central point if there are no external torques. This applies to projectile motion when considering the Earth's center as the origin.

Angular momentum, denoted as \( L \), is dependent on the object's velocity, its mass, and the radial distance from the central point (Earth's center in this case). It is given by:

• \( L = mvr \)

Where:

• \( m \) is the mass of the particle
• \( v \) is the particle's velocity
• \( r \) is the radial distance from the center of rotation

In projectile situations, angular momentum helps us predict the motion path around a central axis, ensuring that the projectile's path and velocity changes align under physics laws without ignoring rotational kinetic components.

Initial Velocity Components

Breaking down initial velocity into its components is crucial for analyzing projectile motion. A projectile's velocity can be divided into horizontal and vertical components, which behave independently of each other.

The horizontal component, \( V_{0x} \), determines the range or distance the projectile will travel horizontally. Meanwhile, the vertical component, \( V_{0y} \), affects how high the projectile will go and how long it stays in the air. These components can be calculated from the initial velocity \( V_0 \) and the angle \( \theta \) of projection as:

• \( V_{0x} = V_0 \cos \theta \)
• \( V_{0y} = V_0 \sin \theta \)

This separation is essential not only for understanding projectile trajectories but also for calculating maximum heights and landing points. By knowing both components, one can predict the entire path of the projectile in a controlled system.

Maximum Height in Projectile Motion

The maximum height reached by a projectile is a critical aspect of its motion, dependent largely on the vertical component of the initial velocity. At this point, the vertical velocity component becomes zero (\( V_y = 0 \)).

Using principles of energy conservation, the maximum height \( h \) can be determined by equating the initial vertical kinetic energy to the gravitational potential energy at the peak. This calculation is concise as:

• \( h = \frac{V_{0y}^2}{2g} \)

Substituting the expression for \( V_{0y} \), we find:

• \( h = \frac{(V_0 \sin \theta)^2}{2g} \)

Knowing the maximum height is useful for applications ranging from sports physics to engineering projectiles or flight paths, ensuring projects are designed to maintain trajectories within safe or required limits.