Q 11-03-096JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
A projectile is projected at $30^\circ$ from horizontal with initial velocity $40\ \text{m s}^{-1}$. The velocity of the projectile at $t=2$ s from the start will be
Answer: (D) $20\sqrt3\ \text{m s}^{-1}$
$v_x=40\cos30^\circ=20\sqrt3$, $v_y=40\sin30^\circ-10\times2=0$. So $v=20\sqrt3\ \text{m s}^{-1}$ (the top of the path, taking $g=10\ \text{m s}^{-2}$).
Solution by Sreeraj P, M.Sc Physics