A

Step 1: Calculate the time of flight for the first shell.
The vertical component of velocity, \( V_{y1} = 40 \sin(60^\circ) = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \, \text{m/s} \).
The time of flight \( T_1 \) is given by \( T = \frac{2V_{y1}}{g} = \frac{2(20\sqrt{3})}{10} = 4\sqrt{3} \, \text{s} \).
Step 2: Calculate the horizontal range of the first shell.
The horizontal component of velocity, \( V_{x1} = 40 \cos(60^\circ) = 40 \times \frac{1}{2} = 20 \, \text{m/s} \).
The range is given by \( R_1 = V_{x1} \times T_1 = 20 \times 4\sqrt{3} = 80\sqrt{3} \, \text{m} \).
Step 3: Calculate the height of point A when the second shell is fired.
The rise of the target is given by the height it reaches before the second shell hits.
The vertical speed is given as \( u = 9\sqrt{3} \, \text{m/s} \).
The height after time \( t \) is given by \( h = ut = 9\sqrt{3} \cdot t \).
Step 4: Set the equation for the second shell.
For the second shell, \( v \sin(60^\circ) \) must equal the height at time of flight.
Therefore, \( u t = \frac{v \sin(60^\circ) t}{g} \) and solves for v.
Therefore, equating the heights: \( 20\sqrt{3}t = 9\sqrt{3}t + v \sin(60^\circ) \cdot t \).
Step 5: Simplifying gives \( v \cdot \frac{\sqrt{3}}{2} = (20 - 9)t \).
This leads to \( v = \frac{11 \cdot 2}{\sqrt{3}} \approx 22 \, m/s \).
Step 6: Therefore, the required speed v is approximately \( v = 22 \, \text{m/s} \). Hence, Option A is the correct answer.