# A stone is falling from the top of a vertical tower. When it has fallen 'a' meter, another stone let dropped from a point 'x' meter below ..?

A stone is falling from the top of a vertical tower. When it has fallen 'a' meter, another stone is dropped from a point 'x' meter below the top. If they fall from rest and reach the ground together, find height of the tower.

Relevance
• Both should have the same velocity at a.

• The first stone, after falling AB ( = a ) m has acquired a velocity of  √(2 ag) and is at 'B' . The second stone now starts from rest from 'C' such that AC = x. and CB  =  (x - a). They reach together at D say in time (t) second. Then --

Distance covered by First stone from B to the ground (BD)----

BD = (h-a)  =  √(2 ag) * t + (1/2) g t² ...... (1)

For the second stone, distance covered (CD) = (h-x)

( h-x ) = 0 t + (1/2) g t² .................. (2)

from (2) :  (1/2) gt²  =  ( h - x ).

Substitute this value in (1)----

(h-a) = √(2 ag) * t + ( h - x )

=> - a  =  √(2 ag) * t  - x

=> t  =  ( x - a )/√(2 ag)  =====>>>  t²  =  (x-a)²/ 2 ag

Substitute this value in (2)

=>  h - x  =  (1/2) g * [ (x-a)²/ 2 ag ]

........    (x-a)² + 4 a x     ..... (x + a)²

=> h  = ------------------  =  ---------------   .......... Answer

................    4 a  ........  ... ..  4 a 