A

Step 1: Let's denote the initial speed of the projectile as \( v_0 = 50 \, \text{m/s} \) and the height of the cliff as \( h = 40 \, \text{m} \).

Step 2: We can solve this problem using conservation of energy. The total mechanical energy at the top will equal the total mechanical energy just before it hits the ground.

The potential energy (PE) at the cliff's top is given by:
\( PE = mgh = mg(40) \)
The kinetic energy (KE) at the top is:
\( KE = \frac{1}{2} mv_0^2 = \frac{1}{2} m(50^2) = \frac{1}{2} m(2500) \)

The total energy at the top is:
\( E_{top} = PE + KE = mg(40) + \frac{1}{2} m(2500) \)

Step 3: Just before hitting the ground, the potential energy will be zero and the kinetic energy at that point will be:
\( KE_{ground} = \frac{1}{2} mv^2 \) where \( v \) is the speed right before impact.

Step 4: By the conservation of energy:
\( E_{top} = KE_{ground} \)
\( mg(40) + \frac{1}{2} m(2500) = \frac{1}{2} mv^2 \)

Step 5: Cancel the mass (m) from both sides:
\( g(40) + \frac{1}{2} (2500) = \frac{1}{2} v^2 \)

Substituting \( g = 10 \, \text{m/s}^2 \):
\( 10(40) + 1250 = \frac{1}{2} v^2 \)
\( 400 + 1250 = \frac{1}{2} v^2 \)
\( 1650 = \frac{1}{2} v^2 \)

Step 6: Multiplying both sides by 2 gives:
\( 3300 = v^2 \)

Step 7: Taking the positive square root:
\( v = \sqrt{3300} \approx 57.45 \, \text{m/s} \)

Therefore, the speed of the projectile just before it hits the ground is approximately \( 57.45 \, \text{m/s} \). Since it matches the speed requirement, option A is confirmed.