Initial angle of a projectile is \(\theta\) and its initial velocity is \(\mathrm{V}_{0}\). Find the angle of velocity with horizontal line at time \(\mathrm{t}\). (A) $\sin ^{-1}\left[1-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (B) $\tan ^{-1}\left[1-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (C) $\tan ^{-1}\left[\tan \theta-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$ (D) $\sin ^{-1}\left[\tan \theta-\left\\{g /\left(\mathrm{V}_{0} \cos \theta\right)\right\\} \mathrm{t}\right]$

The initial horizontal velocity (\(V_{0x}\)) and initial vertical velocity (\(V_{0y}\)) can be found by using trigonometry: \(V_{0x} = V_0 \cos\theta\) \(V_{0y} = V_0 \sin\theta\)

The horizontal velocity does not change with time, as there is no horizontal acceleration, so: \(V_{x} = V_0 \cos\theta\) For the vertical velocity, the acceleration due to gravity acts against the motion. Thus, we can use the following equation of motion to find the vertical velocity at time t: \(V_{y} = V_{0y} - gt\) Substituting the initial vertical velocity from Step 1, we get: \(V_{y} = V_0 \sin\theta - gt\)

The angle of velocity with the horizontal line at time t can be found using the arctangent function: \(\alpha = \tan^{-1}\frac{V_y}{V_x}\) Substituting the horizontal and vertical velocities we found in Step 2, we get: \(\alpha = \tan^{-1}\frac{V_0 \sin\theta - gt}{V_0 \cos\theta}\) Now, we modify the equation by dividing both the numerator and the denominator by \(V_0 \cos\theta\): \(\alpha = \tan^{-1}\left[\frac{\sin\theta - \left(\frac{g}{V_0 \cos\theta}\right)t}{1}\right]\) \(\alpha = \tan^{-1}\left[\sin\theta - \left\{\frac{g}{\left(V_{0} \cos\theta\right)}\right\}t\right]\) Comparing this expression with the given options, we find that it matches option (C). Therefore, the angle of velocity with the horizontal line at time t is given by: \(\alpha = \tan^{-1}\left[\tan\theta-\left\{g /\left(V_{0} \cos\theta\right)\right\}t\right]\)