Two astronauts, one of mass 60 kg and the other 90 kg , are initially at rest together in outer space. they then push each other apart. how far apart are they when the lighter astronaut has moved 15 m ?

Let the mutual force between the two astronauts when they push apart be
 F N, as shown in the figure below.

1st astronaut:
The mass is 60 kg.
Let the acceleration be a₁ m/s².
Then
F = 60a₁            (1)

2nd astronaut:
The mass is 90kg.
The acceleration, a₂, is given by
F = 90a₂           (2)

When lighter astronaut has traveled a distance of 15 m, the time taken, t, is given by
(1/2)a₁t² = 15
or
a₁ t² = 30          (3)

The distance traveled by the other astronaut is
d₂ = (1/2)a₂ t²      (4)

Substitute (3) into (4)
d₂ = (1/2) a₂ (30/a₁) = 15 (a₂/a₁) 

The separation between the astronauts is
D = 15 + d₂ = 15 + 15(a₂/a₁)       (5)

From (1) and (2), obtain
60a₁ = 90a₂
a₂/a₁ = 2/3                       (6)

Substitute (6) into (5).
D = 15 + 15(2/3) = 25 m

25 m

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward