A stone is dropped from a tower 400m high. Simultaneously, another stone is thrown upwards from the earth's surface with a velocity of 100m/s. When and where would these two stones meet? (g=`9.8 m//s^(2))`.

Explanation

Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards.
`therefore` The acceleration of one relative to the other =g-g=0.
At time t=0 , the downward velocity of the two stones are, respectively, 0 and `-100 m//s`. So, relative velocity =0-(-100)=100 m/s. The stones move with this relative velocity , which is uniform in absence of any relative acceleration.
The initial distance between the stones=400m.
Thus, they will meet after a time t=`(400)/(100)=4 s. `
The height through which the first stone falls in time t=4 s is,
`h=1/2"gt"^2=1/2xx9.8xx4^2=78.4m`
`therefore` The stones meet after 4 s at a height of (400-78.4) , or, 321.6m.