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. $ta{n}^{-1}(\frac{\theta }{t})$
• C. $ta{n}^{-1}\left(\frac{v\cos \theta }{v\sin \theta -gt}\right)$
• D. $ta{n}^{-1}\left(\frac{vsin\theta -gt}{v\cos }\right)$

Answer 1

Answer: (D)

According to figure,

$v^{\prime }cos\alpha =vcos\theta$

and $v^{\prime }sin\alpha =vsin\theta -gt$

$\therefore \frac{v^{\prime }sin\alpha }{v^{\prime }cos\alpha }=\frac{vsin\theta -gt}{vcos\theta }$

or tan $\alpha =\frac{vsin\theta -gt}{vcos\theta }$

$\therefore \alpha =ta{n}^{-1}(\frac{vsin\theta -gt}{vcos\theta })$