Question

Two balls are projected upward simultaneously with speeds 40m/s and 60m/s.relative position (x) of second ball w.r.t first ball at time t=5s is

Answer 1

Given:

speed of first ball=40m/s

second ball=60 m/s

time=5s

Distance covered by first ball

S=ut+1/2at²

S=40x5-1/2x10 x 5 x5

s=200-125=75m

Distance travelled by second ball:

S=60x5-1/2x10x5x5
=300-125=175m

Relative position of second ball with respect to first=175-75=100m

Answer 2

So,
Initial velocity u=40m/s
and 60m/s
Also since angle of projection is not mentioned so i am considering that the ball was projected vertically upward
Now
s=ut + 1/2 at²
Where s is the distance covered ,u is the initial velocity  
a is acceleration and t is time taken
So,
a=-9.8 m/s²
But I am considering 10 for easy calculations and also it won't affect the answer since relative position would come out the same
s=ut+ 1/2 at²
s=40×5  + 1/2 × (-10) × 5²
s=75 m
When velocity is 60 m/s
s= 60×5 + 1/2 ×10 × 5²
s=175m
So,
The relative distance  will be 175-75=100m