Question

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

Answer 1

Answer: (D)

velocity at any time t, during the motion of a projectile is given as
$\overset{ \rightarrow }{v}=vcos \, \theta \hat{i}+\left(v sin \theta - g t\right)\hat{j}$
$\therefore $ $tan\alpha =\frac{v_{y}}{v_{x}}=\frac{v \, s i n \theta - g t}{v \, c o s \theta }$