Ball-1 is dropped from the top of a building from rest . At the same moment , ball-2 is thrown upward towards ball-1 with a speed 14 m/s from a point 21 m below the top of building . How far will the ball -1 have dropped when it passed ball-2 (Assume `g=10 m//s^2`)

Options

A. `45/4`m
B. `52/6` m
C. `37/2` m
D. `25/2` m

Answer

A

Explanation

Given ,
Ball-1 is dropped from the top of a building from rest .
Ball-2 is thrown upward towards ball-1 from a point 21 m below the top of building.
Speed of ball - 2= 14 m/s
`g=10 m//s^2`

For ball-1 ,
`h=ut+1/2at^2`
Here, u=0 a=g=10
`rArr h=0+1/2xx10t^4`
`rArr h=5t^2`...(1)
For ball-2 ,
h=21-h , u=14 m/s , a=-g =-10
`rArr` 21-h=14t+`1/2 (-10)t^2`
`rArr 21-h =14t - 5t^2`
21-h=14t -h [ `because ` From equation (1) ]
21=14t
`t=21/14`
`rArr t=3/2`
Then equation (1) becomes ,
`h=5(3/2)^2`
`=5xx9/4`
`=45/4`
`therefore ` Ball-1 have dropped to `45/4` m when it passes ball-2 .