Question

a body is dropped from the top of a tower 100m high simultaneously another ball is thrown upward with a speed of 50 metre per second after what time do they cross each other ​

Answer 1

Suppose they will cross each other after time t then the sum of the value of their displacement should be equal to height of tower.
Now displacement of first ball in time t is gt
2
/2
and for second ball it will be ut−gt
2
/2
adding them we will get ut which should be equal to 100meter
so t=2second.

Answer 2

Answer:

Explanat⇒ the time taken by both balls at point P has same time ' t ' seconds .

⇒ for first ball , that is in the diagram from A to P  

S = ut + 1/2 at²  

we know that u = 0 and  a = g  

⇒ ( 100 - x ) = (0)t + 1/2 gt²------------(1)

now, for the second ball from B to P in the diagram ,  

S = t + 1/2 at²

where u = 50 m/s and a = -g  

therefore, x = 50(t) - 1/2gt² ------------------(2)

by adding equation (1) and (2) we will get ,

100 = 50 t  

implies, t = 100 / 50  

therefore t = 2 sec .

therefore the time that they cross each other is 2 seconds.

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