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

v_y = u_y + a_y t = u sin q – gt

Let angle with horizontal = a tan a

= $\frac { v _ { y } } { u _ { x } }$ = $\frac { u \sin \theta - g t } { u \cos \theta }$

a = tan^–1 $\left( \frac { u \sin \theta - g t } { u \cos \theta } \right)$