Taking vertical upward motion of the first stone for time t, we have
x0=200 m,u=15 m∕ s;
a=−10 m∕ s2,t=t;x=x1
As, x=x0+ut+

1

2

at2
∴‌x1=200+15t+

1

2

(−10)t2
or ‌x1=200+15t−5t2 ....(i)
Taking vertical upward motion of the second stone for time t, we have
x0=200 m,u=30 m s−1,
a=−10 m s−2,t=t,x=x2
Then x2=200+30t−

1

2

×10t2
=200+30t−5t2‌....(ii)
When the first stone hits the ground, x1=0, so
t2−3t−40=0
or ‌(t−8)(t+5)=0
Either t=8 s or −5 s
Since t=0 corresponds to the instant, when the stone was projected.Hence negative time has no meaning in this case. So t=8s. When thesecond stone hits the ground, x2=0, so
0=200+30t−5t2
or t2−6t−40=0
or (t−10)(t+4)=0
Therefore, either t=10s or t=–4s
Since t=–4s is meaningless, so t=10s
Relative position of second stone w.r.t. first is
=x2−x1=15t‌....(iii)
From (i) and (ii)
Since (x2–x1) and t are linearly related, therefore, the graph is a straight linetill t=8s.
For maximum separation, t = 8 s, so maximum separation =15×8=120m after 8 second. Only second stone would be in motion for 2 seconds, sothe graph is in accordance with the quadratic equation, x2=200+30t–5t2for the interval of time 8 seconds to 10 seconds.