The correct option is A 4 s
Ordinary method: Using kinematics:
Let the distance from top of the tower where the stones meet be x,
For first stone taking downward as positive we have from the equation of motion
y=ut+12gt2
Here, y=x and u=0
So, for the first stone the point where the two stones meet is at
x=12gt2(1)
Fo the second stone if we take the upward direction to be positive we have
y=ut+12gt2
Here y=100−x and u=25 m/s
For the second stone the point where the two stones meet is at
100−x=25t−12gt2(2)
Adding (1) and (2), we get
25t=100 or t=4 s
Alternative method: Using Relative velocity
Initial velocity of stone 1 which is dropped from the top of a tower, u1=0
Initial velocity of stone 2 which is thrown upward from the ground, u2=+25 m/s
(Take upward direction to be positive)
Hence,
urel=u21=u2−u1=25−0=25 m/s
srel=100 m
arel=g−g=0
Using second equation of motion, we get
Srel=urelt+12arelt2
⇒100=25×t+0
∴t=4 s
Hence, the two stones will meet after 4 seconds.