Question

A stone is allowed to fall
from the top of a tower 500
m high and at the same time
another stone is projected
vertically upwards from the
ground with a velocity of 50 m/
s. Calculate when and where
the two stones will meet

Answer 1

Answer:

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.