From the top of a tower of height $78.4 m$ two stones are projected horizontally with $10 m / s$ and $20 m / s$ in opposite directions. On reaching the ground, their separation is

120 m

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100 m

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200 m

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150 m

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Solution

The correct option is B $120 m$
For $I^{s t}$ stone $u_{h} = 10 m / s$
$u_{v} = 0; s_{v} = - 78.4 m; g = - 9.8 m / s^{2}$
$s_{v} = u_{v} t + \frac{1}{2} a t^{2}$
$\Rightarrow - 78.4 = \frac{- 9.8}{2} \times t^{2}$
$t^{2} = 16 s e c . \Rightarrow t = 4 s e c .$
Distance travelled in horizontal direction in $4$ s
$s_{n} = 10 \times 4 = 40 m$
For $I I^{n d}$ stone $\to u_{h} = 20 m / s; u_{v} = 0; s_{v} = - 78.4 m; g = - 9.8 m / s^{2}$
It will also take same time as the vertical distance travelled is same. Distance travelled in horizontal direction by $I I^{n d}$ stone in $4$ s
$s_{n} = 20 \times 48 = 80 m$
total distance between stones = $40 + 80 = 120 m$

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From the top of a tower of height 78.4 m two stones are projected horizontally with 10 m/s and 20 m/s in opposite directions. On reaching the ground, their separation is

From the top of a tower of height $78.4 m$ two stones are projected horizontally with $10 m / s$ and $20 m / s$ in opposite directions. On reaching the ground, their separation is