Question

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

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

Answer 1

Answer: (D)

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

Answer 2

tan a = $\frac { \mathrm { v } _ { \mathrm { y } } } { \mathrm { v } _ { \mathrm { x } } } = \frac { \mathrm { v } \sin \theta - \mathrm { gt } } { \mathrm { v } \cos \theta }$