Ball-1 is dropped from the top of a building from rest. At the same moment, ball-2 is throuwn upward toward 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 passes ball-2. (Assume acceleration due to gravity, `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

Suppose that both the balls meet at a distance h from ball-1 after time t from the start as shown in the figure.
For downward motion of ball-1 from second equation of the motion,
`h=u t+1/2 g t^(2)` ...(i)
`h=0xxt+1/2 xx10 t^(2)" "[ :' u=0]`
`h=5t^(2)` ...(ii)
For upward motion of ball-2, from second equation of the motion,
`h=v t-1/2 g t^(2)` ...(iii)
Given, speed of ball-2, `v=14` m/s, acceleration due to gravity, `g=10 m//s^(2)`
Substituting these values in Eq. (iii), we get
`21-h=14 t-1/2 xx10 t^(2)`
`implies 21-h=14t-5 t^(2)` ...(iv)
Adding Eq. (ii) and (iv), we get
`21=14 t`
`implies t=3/2 s=1.5 s`
`:.` From Eq. (ii), we get
`h=5xx(1.5)^(2)=11.25 m`
`=45/4 m`
Therefore, the ball-1 will have dropped `45/4` m when it passes ball-2.