A body is falling freely from a point A at a certain height from the ground and passes through points B,C and D vertically as shown below so that BC = CD. The time taken by the particle to move from B to C is 2 s and from C to D is 1s. Time taken to move from A to B in seconds is

Options

A. 0.6
B. 0.5
C. 0.2
D. 0.4

Answer

b

Explanation

Let velocity of the particle at point B be V.
Now `RC = 2V + (1)/(2) 0 xx (2)^(2)`
or, `BC = 2V + 2g`
similarly `28C = 3V + (1)/(2) xx g xx (3)^(2)`
`implies 2BC = 8v+ (9g)/(2)`
From Eq. (i)
`2 (2V + 2g) = 3V + (9g)/(2)`
`implies 4v + 4g = 3v + (9g)/(2)`
`implies 4v - 3v = (9g)/(2) - 4g`
`implies v = (g)/(2)`
From point A to B
v = u + at
`implies `(g)/(2) = 0 + g xx t`
`implies t (1)/(2)` second
`implies t = 0.5 s`