Question
Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by, $\theta(\text{t})=\tan^{-1}\Big(\frac{\text{u}_\text{oy}-\text{gt}}{\text{u}_\text{ox}}\Big)$
✓
Answer
Let $v_{ox}$ and $v_{0y}$ respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions. Let $v_x$ and $v_y$ respectively be the horizontal and vertical components of velocity at a point P. 
Time taken by the projectile to reach point P = t Applying the first equation of motion along the vertical and horizontal directions, we get: $v_y = v_{oy} = gt$ And $v_x = v_{ox}$
$\therefore\tan\theta=\frac{\text{v}_\text{y}}{\text{v}_\text{x}}=\frac{\text{v}_\text{oy}-\text{gt}}{\text{v}_\text{ox}}$ $\theta=\frac{\tan^{-1}(\text{v}_\text{oy}-\text{gt})}{\text{v}_\text{ox}}$
Time taken by the projectile to reach point P = t Applying the first equation of motion along the vertical and horizontal directions, we get: $v_y = v_{oy} = gt$ And $v_x = v_{ox}$
$\therefore\tan\theta=\frac{\text{v}_\text{y}}{\text{v}_\text{x}}=\frac{\text{v}_\text{oy}-\text{gt}}{\text{v}_\text{ox}}$ $\theta=\frac{\tan^{-1}(\text{v}_\text{oy}-\text{gt})}{\text{v}_\text{ox}}$
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