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Two stones are through up simultaneously...

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)

A

B

C

D

Text Solution

Verified by Experts

The correct Answer is:
B
(b):
For Ball A, time taken to reach the ground is t
`S_(1)=u_(1)t-(1)/(2)"gt"^(2)`
`-240=10t-(1)/(2)xx10t^(2)`
`5t^(2)-10t-240=0`
`t=+8 s, -6 s`
For Ball B, time taken to reach the ground with velocity
`u=40m//s` is `t_(1)`
`S_(2)="ut"_(1)-(1)/(2) a t_(1)^(2)`
`-240=40t_(1)-(1)/(2) 10 t_(1)^(2)`
`5t_(1)^(2)-40t_(1)-240=0`
`t_(1)=12` seconds.
During first 8 seconds .
`S_(1)-S_(2)=(u_(1)t-u_(2)t)`
`S_(1)-S_(2)=(u_(1)-u_(2))t`
`S_(1)-S_(2)= -30t`
`rArr (S_(2)-S_(1))=30t`
This is straight line for 8 seconds, for next four seconds only ball B is in air .
Hence for next 4 s it should be parabolic curve , as shown in option (b) .
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