Question

A stone is dropped from the top of a building 20m high at the same instant another stone is projected up from the ground with velocity of 20m/s when and where two stones will meet

Answer 1

there are two cases :
case 1 :- A stone is dropped from the top of a building 20m hight.
initial velocity of ball, u = 0
distance S from the ground where the stone is collided with another stone.
use formula , $S=S_0+ut+\frac{1}{2}at^2$
S = 20 - 1/2 × 10 × t²
S = 20 - 5t² ........(1)


case 2 :- a stone is thrown from ground with speed 20m/s , after t sec it strikes with 1st stone.
so, distance , S'= 20t - 1/2 × 10t²
= 20t - 5t² .........(2)

from equation (1) and (2),
S = S'
20 - 5t² = 20t - 5t²
t = 1 sec

S = 20 - 5 = 15 m
hence, 15m above from the ground both stone are collided.