JEE PYQ: Motion In Two Dimensions Question 22
Question 22 - 2019 (11 Jan Shift 1)
A body is projected at $t = 0$ with a velocity $10$ ms$^{-1}$ at an angle of $60°$ with the horizontal. The radius of curvature of its trajectory at $t = 1$ s is R. Neglecting air resistance and taking acceleration due to gravity $g = 10$ ms$^{-2}$, the value of R is:
(a) 10.3 m (b) 2.8 m (c) 2.5 m (d) 5.1 m
Show Answer
Answer: (b)
Solution
At $t = 1$: $v_x = 5$ m/s, $v_y = 5\sqrt{3} - 10$ m/s. $\theta = 15°$. $R = 100(2 - \sqrt{3})/(10\cos 15°) \approx 2.8$ m.
BETA
