A man of mass $60 k g$ start falling from building of height $80 m$ with a bag of $2 k g$ in his hand. After falling through a distance $20 m$ he throw the bag horizontally with respect to him so that he fall in a pond $2 m$ away from the vertical lines of fall. Calculate horizontal distance of bag from the vertical line of fall where it lands (take $g = 10 m / s^{2})$

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Solution

By using

$S = u t + \frac{1}{2} a t^{2}$

Time taken to cover $2 m$ distance is $2 sec$

Time taken to cover rest of the height is $2 sec$

The velocity of man in horizontal direction is

$v_{H M} = \frac{2 m}{2 sec}$

$v_{H M} = 1 m / s$

Using the conservation of momentum we get

$60 \times 1 = 2 \times v_{b}$

The horizontal distance of a bag from vertical line is

$s = 30 \times 2$

$s = 60 m$

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By using S=ut+12at2 Time taken to cover2mdistance is2sec Time taken to cover rest of the height is2sec The velocity of man in horizontal direction is vHM=2m2sec vHM=1m/s Using the conservation of momentum we get 60×1=2×vb The horizontal distance of a bag from vertical line is s=30×2 s=60m