To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.
Gravitational potential energy can be defined as

As M=m, then

Where,
m = Mass
G =Gravitational Universal Constant
R = Distance /Radius
PART A) As half its initial value is u'=2u, then



Therefore replacing we have that,

Re-arrange to find v,



Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s
PART B) With a final separation distance of 2r, we have that

Therefore




Therefore the velocity when they are about to collide is 
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Answer:
S = V t where S is the horizontal distance traveled
1/2 g t^2 = H where H is the vertical distance traveled
t^2 = 2 H / g
V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations
tan theta = H / S
V^2 = S g / (2 tan theta)
Using S = L cos theta
V^2 = L g cos theta / (2 tan theta)
Giving V in terms of L and theta