Two particles A and B are thrown vertically upward with velocity, $5 m / s$ and $10 m / s$ respectively ( $g = 10 m / s^{2}$ ). Find separation between them after one second.

1 m

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2 m

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5 m

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10 m

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Solution

The correct option is C
5 m

Position of A after 1 s, $s_{A} = u t - \frac{1}{2} g t^{2} = 5 \times 1 - \frac{1}{2} \times 10 \times 1^{2} = 0$
i.e., the particle will return to ground at t = 1 s.
The position of B after 1 s, $s_{B} = u t - \frac{1}{2} g t^{2} = 10 \times 1 - \frac{1}{2} \times 10 \times 1^{2} = 5 m$
Hence, separation between A and B, $s_{B} - s_{A} = 5 m$

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