A ball is dropped from the top of a tower 100m high simultaneously another ball is thorwn upward with a speed of 50 meter per second. After what time do they cross each other?
(a) 1s
(b) 2s
(c) 3s
(4) 4s

Question

A ball is dropped from the top of a tower 100m high simultaneously another ball is thorwn upward with a speed of 50 meter per second. After what time do they cross each other?
(a) 1s
(b) 2s
(c) 3s
(4) 4s

anonymous · Answer

When they cross each other they will have the same displacement. Set up equations for the displacements of each and equate them.

anonymous · Answer

but which equation, i've tried by appling S=vi t+1/2 at^2 but that does not work for me.Please can you solve it?

anonymous · Answer

You are on the right track. For the ball dropped downward, vi = 0. 
So$$\color{green}{\vec{d} = \frac{1}{2}\vec{g}t^2 }$$
Set up the other equation and equate the two and solve for t.

anonymous · Answer

i've tried many methods but couldn't solve it,please solve it if you can

pawanyadav · Answer

Assume displacemt x  and   100-x  for thrown  and release one.
Then
           100-x=0.5gt^2
            x=100-5t^2
Again 
             For the thrown ball
        x=50t-5t^2
As  time for both balls are equal
  Equating both values you will get the answer 2.

shamim · Answer

Did u get it?

shamim · Answer

We wanna help u. Response plz!

anonymous · Answer

ok,thanks i've got it