1. Initial velocity components

Given: $u = 10\ \text{m s}^{-1}, \quad \theta = 60^\circ, \quad g = 10\ \text{m s}^{-2}$

Resolve the initial velocity: $u_x = u\cos 60^\circ = 10\cdot \frac{1}{2} = 5\ \text{m s}^{-1}$ $u_y = u\sin 60^\circ = 10\cdot \frac{\sqrt{3}}{2} = 5\sqrt{3}\ \text{m s}^{-1}$

2. Velocity at $t=1\,$s

Horizontal velocity remains constant: $v_x = 5\ \text{m s}^{-1}$

Vertical velocity after 1 s: $v_y = u_y - gt = 5\sqrt{3} - 10$

So speed at $t=1$ s is $v = \sqrt{v_x^2 + v_y^2}$ $v^2 = 5^2 + (5\sqrt{3}-10)^2$ $= 25 + 25\cdot 3 + 100 - 100\sqrt{3}$ $= 200 - 100\sqrt{3}$

Using $\sqrt{3} \approx 1.732$, $v^2 \approx 200 - 173.2 = 26.8$ $v \approx 5.18\ \text{m s}^{-1}$

3. Radius of curvature formula

For planar motion, $R = \frac{v^3}{|\vec v \times \vec a|}$

Here acceleration is $\vec a = (0,-10)$

Velocity is $\vec v = (5,\ 5\sqrt{3}-10)$

Magnitude of cross product: $|\vec v \times \vec a| = |5(-10) - (5\sqrt{3}-10)(0)| = 50$

Hence, $R = \frac{v^3}{50}$

Now, $v^3 = v\cdot v^2 \approx 5.18\times 26.8 \approx 138.8$

Therefore, $R \approx \frac{138.8}{50} \approx 2.78\ \text{m}$

4. Final answer

$R \approx 2.8\ \text{m}$

So the correct option is A.