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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:
C
For the second stone time required to reach the ground is given by
` y = ut -1/2 "gt"^(2)`
` -240 = 40t -1/2 xx 10 xx t^(2)`
`(t-12) (t+8)=0`
` :. T=12s `
For the first stone :
` - 240 = 10t -1/2 xx 10 xx t^(2)`
` :. - 240 = 10t - 5t^(2)`
`5t^(2) - 10 t - 240 = 0`
`(t-8)(t+6)=0`
T=8s
During first 8 seconds both the stones are in air :
` :. y_(2) -y_(1) =(u_(2)-u_(1))t=30 t`
` :. ` graph of `(y_(2)-y_(1))` against t is a straight line
After 8 seconds
`y_(2)=u_(2)t-1/2"gt"^(2) - 240`
Stone two has acceleration with respect to stone one .
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