Two stones are through 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 graphs best represents the time variation of relative position of the second stone with respect to the first? Assume stones do not rebound after hitting the ground and neglect air resistance, take . ` g= 10 m//s^(2) ` (The figures are schematic and not drawn to scale)

Options

A.
B.
C.
D.

Answer

a

Explanation

Initial speed of stone `1,u_(1)=10m//s`
Initial speed of stone-2, `u_(2)=40m//s`

Let us first calculate the time taken by each stone to reach to the ground.
Time taken by stone-1 to reach the ground
`-240=10t-(1)/(2)xx10t^(2)`
`5t^(2)-10t-240=0rArr t^(2)-2t-48=0`
Thus t=8s
Time taken by stone 2 reach the ground
`-240=40t-(1)/(2)xx10t^(2)`
`5t^(2)40t-240=0rArrt^(2)-8t-48=0`
t=12,-4
Thus, t=12s
Let `y_(1)` be the instantaneous position of the stone-1 at time t:
`y_(1)=10t-5t^(2)`
Let `y_(2)` be the instantaneous position of the stone-2 at time t, then its displacement can be written as,
`y_(2)=40t-5t^(2)`
Thus, the difference of the position is given by:
`y_(2)-y_(1)=30t`
For time `8s le t le 12s`
`y_(2)=40t-5t^(2)`
Thus, the difference of the position is given by:
`y_(2)-y_(1)=-5t^(2)+40t+240`
Hence, the correct schematic diagram is as given in option (a).