Question

A stone is thrown vertically upwards with such a velocity as will just take it to the level of
the top of a tower 100m high. Two seconds later, another stone is thrown up from the same
place with the same velocity. Find when and where the stones will meet.

Answer 1

Explanation:

Let the stones meet at point A after time t.

For upper stone  :            

u′=0         

x=0+21gt2

x=21×10×t2           

⟹x=5t2            ............(1)

For lower stone :            

u=25  m/s  

100−x=ut−21gt2

100−x=(25)t−21×10×t2           

⟹100−x=25t−5t2            ............(2)

Adding (1) and (2),  we get          

25t=100              

⟹t=4 s        

From (1),            

x=5×42                 

⟹x=80 m 

Hence the stone meet at a height of   20 m above the ground after 4 seconds.

Answer 2

Answer:

The stones will meet at height 80m