∴−5t2+15t+200=0On solving we can get:
t=8s or
t=–5sSince the stone was projected at time
t=0, the negative sign before time is meaningless.
∴t=8sFor second stone:
Initial velocity, u2=30m/s
Acceleration, a=−g=−10m/s2
Using the relation,
x2=xo+u2t+12at2
=200+30t−5t2 ........(ii)
At the moment when this stone hits the ground; x2=0
−5t2+30t+200=0
t=10s or t=−4s
Here again, the negative sign is meaningless.
∴t=10s
Subtracting equations (i) and (ii), we get
x2−x1=(200+30t−5t2)−(200+15t−5t2)
x2−x1=15t........(iii)
Equation (iii) represents the linear path of both stones. Due to this linear relation between (x2−x1) and t, the path remains a straight line till 8s.
Maximum separation between the two stones is at t=8s
(x2−x1)max=15×8=120m
This is in accordance with the given graph.
After 8s, only second stone is in motion whose variation with time is given by the quadratic equation:
x2−x1=200+30t−5t2
Hence, the equation of linear and curved path is given by
x2−x1=15t (Linear path)
x2−x1=200+30t−5t2 (Curved path)