Two stones are thrown up simultaneously from the edge of a cliff ` 240 m ` high with initial speed of ` 10 m//s and 40 m//s` respectively . Which of the following graph best represents the time variation of relative position of the speed stone with respect to the first ?
( Assume stones do not rebound after hitting the groumd and neglect air resistance , take ` g = 10 m//s^(2))`
( The figure are schematic and not drawn to scale )

Options

A. (a)
B. (b)
C. (c )
D. (d) .

Answer

C

Explanation

Taking vertical upward motion of first stone for tiem (t) we have
` y_0 = 240 m, u= 10 m//s, a =- 10 m//s`
` t=t, y =y`
As ` y =y_0 + ut + 1/2 at^2`
:. ` y_1 = 240 + 10t + 1/2 (-10) t^2 = 240 + 10 t- 5 t^`
Taking vertical upward motion of second stone for time (t) we have ` y_0 = 240 , u = 40 m//s`,
` a=- 10 m//s , t= t, y= y_2`
`y _2 = 240 + 40 t- 1/2 xx 10 xx t^2 = 240+ 40 t - 5 t^2`
When first stone hits the ground, ` y_1 =0`, ltbRgt :. `0= 240 + 10 t- 5 t^2` or ` t^2 - 2 48 =0`
pr ` t= ( 2 +- sqrt((2)^2 + 4xx 48 )/2 = 8 s ` or `-6 sec`
`-6` s is meaningless so time ` t= 8 sec`
When second stone hits the ground, ` y_2 =0`
:. ` =0= 240 + 40 t- 5 t^2` or `t^2 - 8 t- 48 =0`
on solving we get, ` t = 12 s ` or ` -4 s`
as `-4` s is meaningless so, ` t= 12 s`
Relatve position of `2 nd` stone w.r.t. first is
` y_2 -y_1 = 30 t...(i) [0 ltt-lt 8 s]` ltbRgt Since ` y_2 -y_1` and (t) are lojearly related.
` graph will be straight line from ` t= 0` to ` t=8 s`.
After ` 8sec` only second stone would be in motion for further ` 4 s`. Now ` y_1 = 240 m` and
` y_2 = 240 + 40 t- - 5 t^2`
:. ` y_2 - y-1 = 40 t- 5 t^2` ....(ii) ` (8 -lt t-lt12 s)` ltbRgt Thes potion (c ) is true.