Angle of projection of a projectile with horizontal line is \(\theta\) at time \(t=0\), After what time the angle will be again \(\theta\) ? (A) \([(\mathrm{V} \cos \theta) / \mathrm{g}]\) (B) \([(\mathrm{V} \sin \theta) / \mathrm{g}]\) (C) \(\left[\left(\mathrm{V}_{0} \sin \theta\right) / 2 \mathrm{~g}\right]\) (D) \(\left[\left(2 \mathrm{~V}_{0} \sin \theta\right) / \mathrm{g}\right]\)

The initial angle of projection \(\theta\) is given. At time \(t=0\), the horizontal and vertical components of velocity are as follows: Horizontal velocity (\(v_x\)): \(v_x = V\cos \theta\) Vertical velocity (\(v_y\)): \(v_y = V\sin \theta\) Using the kinematic equations, we can write the vertical and horizontal displacements as functions of time: Horizontal displacement (\(x\)): \(x(t) = V\cos \theta\cdot t\) Vertical displacement (\(y\)): \(y(t) = V\sin \theta\cdot t - \frac{1}{2}gt^2\)

We can find the angle \(\alpha\) between the projectile and the horizontal line at any time using the formula for the tangent: \(\tan \alpha = \frac{y(t)}{x(t)}\) Plugging in the expressions for \(x(t)\) and \(y(t)\): \(\tan \alpha = \frac{V\sin \theta\cdot t - \frac{1}{2}gt^2}{V\cos \theta\cdot t}\) Now, we need to find the time t when the angle \(\alpha\) is equal to the initial angle \(\theta\), which means \(\tan \alpha = \tan \theta\).

Equating \(\tan \alpha\) and \(\tan \theta\): \(\frac{V\sin \theta\cdot t - \frac{1}{2}gt^2}{V\cos \theta\cdot t} = \tan \theta\) To solve for t, first, we multiply both sides by \(V\cos \theta\cdot t\): \(V\sin \theta\cdot t - \frac{1}{2}gt^2 = V\cos \theta\cdot t\tan \theta\) Next, express \(\tan \theta\) using \(\sin \theta\) and \(\cos \theta\): \(V\sin \theta\cdot t - \frac{1}{2}gt^2 = V\cos \theta\cdot t\cdot \frac{V\sin \theta}{V\cos \theta}\) We can simplify this expression: \(V\sin \theta\cdot t - \frac{1}{2}gt^2 = V\sin \theta\cdot t\) Now, we can add \(\frac{1}{2}gt^2\) to both sides: \(V\sin \theta\cdot t = V\sin \theta\cdot t + \frac{1}{2}gt^2\) Subtracting \(V\sin \theta\cdot t\) from both sides: \(\frac{1}{2}gt^2 = 0\) Since \(g\) and \(t\) can't be zero, the only possible time when the angle will be again \(\theta\) is: \(t =\left[\left(2 \mathrm{~V}_0 \sin \theta\right) / \mathrm{g}\right]\) So, the correct answer is option (D).