A bag is dropped from a helicopter rising vertically at a constant speed of $2$ $m / s$ . The distance between the two after $2$ s is (Given $g = 9.8 m / s^{2}$ )

4.9 m

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19.6 m

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29.4 m

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39.2 m

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Solution

The correct option is B $19.6 m$
for helicopter
$a = 0, t = 2 s e c, u = 2 m / s, h_{1} = u t$
for body
$h_{2} = - u t + \frac{1}{2} g t^{2}$
distances between them $= h_{1} - h_{2}$

So,
Distance travelled by pocket in $2$ sec,
$s_{1} = 2 (2) - \frac{1}{2} g (2)^{2} = 4 - 5 (2)^{2} = - 15.6 m$
Distance travelled by helicopter in $2$ sec
$s_{2} = 2 (2) = 4 m$
Hence, distance between the two $\Rightarrow s_{2} - s_{1} = 4 - (- 15.6)$ $= 19.6 m$

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A bag is dropped from a helicopter rising vertically at a constant speed of 2 m/s. The distance between the two after 2 s is (Given g=9.8m/s2)

A bag is dropped from a helicopter rising vertically at a constant speed of $2$ $m / s$ . The distance between the two after $2$ s is (Given $g = 9.8 m / s^{2}$ )