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 $\left[\left(\mathrm{V}_{0}^{2} \sin ^{2} \theta\right) / 2 \mathrm{~g}\right]$

Step by step solution

01

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

02

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 \(L = mvr\).

03

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)

04

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)

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