The angle which the velocity vector of a projectile, thrown with a velocity $v$, at an angle $\theta$ to the horizontal, will make with the horizontal after time $t$ of its being thrown up is 

• A

  $\mathrm{\theta }$
• B

  ${\tan }^{-1}\left(\frac{\theta }{t}\right)$
• C

  ${\tan }^{-1}\left(\frac{v\mathrm{ }\cos \theta }{v\mathrm{ }\mathrm{sin\theta }-\mathrm{g}t}\right)$
• Correct answer
  D

  ${\tan }^{-1}\left(\frac{v\mathrm{ }\sin \theta -\mathrm{g}t}{v \cos \theta }\right)$

Solution

velocity at any time t, during the motion of a projectile is given as

$\overrightarrow{v}=vcos \theta \widehat{i}+\left(v\sin \theta -gt\right)\widehat{j}$

$\therefore$$\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{\alpha }=\frac{v_{\mathrm{y}}}{v_{\mathrm{x}}}=\frac{v\mathrm{ }\mathrm{s}\mathrm{i}\mathrm{n}\theta -\mathrm{g}t}{v \mathrm{c}\mathrm{o}\mathrm{s}\theta }$
 

Hence, the correct answer is option: 'D'.