Question

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

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

Answer 1

Answer: (D)

According to figure,
$v'\,\,cos\,\,\alpha=v\,\,cos\,\,\theta$
and $v' \,\,sin\,\,\alpha=v\,\,sin\,\,\theta-gt$
$\therefore \frac{v' \,\,sin\,\,\alpha}{v'\,\,cos\,\,\alpha}=\frac{v\,\,sin \,\theta-gt}{v\,\,cos\,\,\theta}$
or tan $\alpha=\frac{v \,\,sin\,\, \theta-gt}{v\,\,cos\,\,\theta}$
$\therefore \alpha=tan^{-1}(\frac{v\,\,sin\theta-gt}{v\,\,cos\,\,\theta})$