## anonymous one year ago 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

1. anonymous

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

2. anonymous

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?

3. anonymous

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.

4. anonymous

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

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.

6. shamim

Did u get it?

7. shamim

We wanna help u. Response plz!

8. anonymous

ok,thanks i've got it