Question

A stone is dropped from top of a tower 100 m height. At the same instant another stone is thrown vertically upward from the base of the tower with the velocity of 25 m/s. When and where will the two stones meet? 

Given g = 10 m/s².

Answer 1

For Stone A, u = 0, S = x and a = + g = 10 m/s² 

Formula s = ut + 1/2 at² 

Arriving at x = 5t² —(1) 

For stone B s = 100 – x, a = – g and u = 25 m/s 

Arriving at 100 – x = 25t – 5t² …(2) 

Obtaining t = 4 second 

Obtaining x = 80 m