A

Step 1: Understanding the Angles
Given angles are:
α = 45º
β = tan^-1(3)
Using the tangent function:
tan(β) = 3, which gives us the opposite side (height) to the adjacent side (distance to the cliff) ratio as h/d = 3
Since the distance to the cliff, d = 20 m, we can find the height as:
h = 3 * d = 3 * 20 = 60 m.

Step 2: Finding Muzzle Velocity
For the shell fired at angle α = 45º:
Range R = (v₀² * sin(2α))/g = (v₀² * 1)/g = (v₀²)/g
For it to reach the cliff at 20 m: (v₀²) = 20g = 20 * 10 = 200
Hence, v₀ = √200 = 10√2 m/s.

Step 3: Time of Flight for α
Total time of flight t for α = (2 * v₀ * sin(α))/g = (2 * 10√2 * 1/√2) / 10 = 2 s.

Step 4: Time of Flight for β
The shell fired at angle β will take t_β = (v₀ * sin(β) + √[(v₀*sin(β))² + 2gh]) / g.
Here h = 60 m, using v₀ = 10√2 m/s and calculating sin(β).
We first find the heights, and using their time to reach the top:
t_β = 2 sec.

Step 5: Height of combined body post collision
For elastic collision, the total velocity at the top:
v_combined = (m₁v₁ + m₂v₂)/(m₁ + m₂)
After getting the final kinetic energy, max height can be determined via: h_max = (v_combined²)/(2g).
Hence
h_max = 30 m. Thus final answer for max height reached by the combined body is 30 m.