Q 11-03-119JEE MainJEE Main 2022 (27 Jun, Shift 1)Medium
A projectile is launched at an angle $\alpha$ with the horizontal with a velocity $20\ \text{m s}^{-1}$. After $10$ s, its inclination with horizontal is $\beta$. The value of $\tan\beta$ will be : $(g = 10\ \text{m s}^{-2})$.
Answer: (B) $\tan\alpha - 5\sec\alpha$
After $t = 10$ s:
$v_x = 20\cos\alpha$ (unchanged), $\quad v_y = 20\sin\alpha - 10(10) = 20\sin\alpha - 100$.
$$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha} = \tan\alpha - 5\sec\alpha$$
Solution by Sreeraj P, M.Sc Physics