A ball is dropped downwards, after $1$ sec another ball is dropped downwards from the same point. What is the distance between them after $3$ sec? (Take $g = 10 m s^{- 2}$ )

20 m

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25 m

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50 m

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9.8 m

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Solution

The correct option is B $25 m$
Distance at a given time is given by equation,
$S = u t + \frac{1}{2} a t^{2}$
For first body, $u_{1} = 0, S = S_{1}, a = g, t_{1} = 3 s e c$
$∴ S_{1} = \frac{1}{2} g \times 9$
For second body, $u_{2} = 0, S = S_{2}, a = g, t_{2} = 2 s e c$
$∴ S_{2} = \frac{1}{2} g \times 4$
So difference between them after $3 s e c = S_{1} - S_{2}$
$= \frac{1}{2} g \times 5$
If $g = 10 m / s^{2}$ then $S_{1} - S_{2} = 25 m$

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A ball is dropped downwards, after 1 sec another ball is dropped downwards from the same point. What is the distance between them after 3 sec? (Take g=10ms−2)

The correct option is B 25mDistance at a given time is given by equation,S=ut+12at2For first body, u1=0, S=S1,a=g, t1=3sec∴S1=12g×9For second body, u2=0, S=S2, a=g, t2=2sec∴S2=12g×4So difference between them after 3sec=S1−S2 =12g×5 If g=10 m/s2 then S1−S2=25m

A ball is dropped downwards, after $1$ sec another ball is dropped downwards from the same point. What is the distance between them after $3$ sec? (Take $g = 10 m s^{- 2}$ )