Question

A stone projected from the foot of a tower of height 100m with a velocity 25m/s, at the same time another stone is dropped from the top of the same tower. Calculate the time after the stones will meet.

Answer 1

Let the stones meet at point A after time t.

For upper stone  :            

u  

′

=0          

x=0+  

2

1

​

gt  

2

 

x=  

2

1

​

×10×t  

2

           

⟹x=5t  

2

            ... ... ... ...(1)

For lower stone :            

u=25  m/s  

100−x=ut−  

2

1

​

gt  

2

 

100−x=(25)t−  

2

1

​

×10×t  

2

           

⟹100−x=25t−5t  

2

            ... ... ... ...(2)

Adding (1) and (2),  we get          

25t=100              

⟹t=4 s        

From (1),            

x=5×4  

2

                 

⟹x=80 m  

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