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.

5 m

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12 m

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3 m

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7 m

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Solution

The correct option is A 5 m
$s_{A} = u t - (1 / 2) g t^{2}$
$= 5 t - \frac{1}{2} \times 10 \times t^{2}$
$= 5 \times 1 - 5 \times 1^{2}$
$= 5 - 5 = 0 m$
$s_{B} = u t - \frac{1}{2} g t^{2}$
$= 10 \times 1 - \frac{1}{2} \times 10 \times 1^{2}$
$= 10 - 5 = 5 m$
Therefore, $s_{B} - s_{A} = s e p a r a t i o n = 5 m$

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