A body is projected at $t = 0$ with a velocity $10\ \text{m s}^{-1}$ at an angle of $60^\circ$ 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 $g = 10\ \text{m s}^{-2}$, the value of $R$ is:
Answer: (B) $2.8$ m
At $t = 1$ s:
$$v_x = 10\cos60^\circ = 5\ \text{m/s},\qquad v_y = 10\sin60^\circ - 10(1) = 8.66 - 10 = -1.34\ \text{m/s}$$
$$v^2 = 25 + 1.80 = 26.8\ \text{m}^2/\text{s}^2,\qquad v = 5.18\ \text{m/s}$$
Only the component of $g$ perpendicular to the velocity bends the path:
$$a_\perp = g\,\frac{v_x}{v} = 10\times\frac{5}{5.18} = 9.66\ \text{m/s}^2$$
$$R = \frac{v^2}{a_\perp} = \frac{26.8}{9.66} \approx 2.8\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics