1.

A person throws ball with velocity from the top of a building in vertically upward direction the ball reaches the ground with a speed of 3v then the height of the building is :

Answer»

`(4v^(2))/(g)`
`(3V^(2))/(g)`
`(6v^(2))/(g)`
`(9V^(2))/(g)`

Solution :Let h be the height and x be the maximum height of the BALL above the building to WHICHIT rise. Then we have
`x=(v^(2))/(2g)` and `h+x=(3v)^(2)/(2g)`
`:. H=(9v^(2))/(2g)-(v^(2))/(2g)=(4v^(2))/(g)`


Discussion

No Comment Found

Related InterviewSolutions