Analysing the question:
since the ball is thrown upwards, the gravity will be applied opposite to the motion of the ball
We are given:
acceleration due to gravity (a) = -10 m/s²
initial upward velocity (u) = 16.6 m/s
final upward velocity (v) = 5.3 m/s
distance between the friends (s) = s m
Neglecting air resistance
Solving for the distance between the friends:
from the first equation of motion
v = u + at
replacing the variables
5.3 = 16.6 + (-10)(t)
-10(t) = 5.2 - 16.6
-10 (t) = -11.3
t = 1.13 seconds
from the second equation of motion
s = ut + 1/2 at²
replacing the variables
s = (16.6)(1.13) + -5(1.13)²
s = 18.8 - 6.4
s = 12.4 m
Therefore, the 2 friends are 12.4 m apart