A ball is dropped from the top of a tower of height 100m. Simultaneous

1.	A ball is dropped from the top of a tower of height 100m. Simultaneously, another ball was thrown of 50 m/s(g 10 m/s2). These two balls would cross each otherafter a time(A) 1 second(C) 3 seconds(B) 2 seconds(D) 4 seconds
Answer» ⇒ 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. ------------------------------------------------------ answer of this question is 2 second

A ball is dropped from the top of a tower of height 100m. Simultaneously, another ball was thrown of 50 m/s(g 10 m/s2). These two balls would cross each otherafter a time(A) 1 second(C) 3 seconds(B) 2 seconds(D) 4 seconds

Answer»

⇒ 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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answer of this question is 2 second

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