The correct option is B Time =4 s ; Height =80 m
The height of the tower, h=100 m
Initial velocity of the second stone, u2=25 m s−1
g=10 m s−2
Initial velocity of the first stone, u1=0 m s−1
Let t be the time instant at which the two stones would meet/hit each other.
For the falling stone, the distance covered:
For the rising stone, the distance covered:
⟹25t=100⟹t=4 s ----- (2)
Substitute (2) in (1):
s1=5×(4)2=5×16=80 m --- (3)
∴ The two stones would meet each other at t=4 s, and at a height of 80 m.