A stone is thrown upwards with a velocity $50 m g^{- 1}$ . Another stone is simultaneously thrown downwards from the same location with a velocity $50 m s^{- 1}$ . When the first stone is at the highest point, the relative velocity of the second stone w.r.t. the first stone is:

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50 m s^{- 1}

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100 m s^{- 1}

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150 m s^{- 1}

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Solution

The correct option is C $100 m s^{- 1}$
Using concept of relativity the relative acceleration $= {\to a}_{1} - {\to a}_{2} = g - g = 0$
And, Relative velocity $= {\to v}_{1} - {\to v}_{2} = 50 - (- 50) = 100 m / s$
Since relative acceleration is zero, relative velocity will be constant.

Therefore, C is correct option.

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A stone is thrown upwards with a velocity 50mg−1. Another stone is simultaneously thrown downwards from the same location with a velocity 50ms−1. When the first stone is at the highest point, the relative velocity of the second stone w.r.t. the first stone is:

The correct option is C 100ms−1Using concept of relativity the relative acceleration =→a1−→a2=g−g=0And, Relative velocity=→v1−→v2=50−(−50)=100m/sSince relative acceleration is zero, relative velocity will be constant.Therefore, C is correct option.

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