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A small body is projected vertically upward from a point on the ground with a velocity of 50 m/s. The body arrives at the ground with a velocity of 40 m/s. If the mass of the body is 1 kg and g $= 10 m / s^{2}$ , the average resistance force offered by air to the motion of the body is

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$\frac{87}{42} N$

$\frac{172}{93} N$

$\frac{90}{41} N$

$\frac{85}{44} N$

answer is C.

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Detailed Solution

If R be the average air drag and H be the maximum height attained by the body, then

$\frac{1}{2} {mu}^{2} = mgH + RH . . . . . . . (1)$

$\frac{1}{2} {mv}^{2} = mgH - RH . . . . . . . (2)$

$∴ \frac{u^{2}}{v^{2}} = \frac{mg + R}{mg - R} \Rightarrow R = \frac{mg (u^{2} - v^{2})}{(u^{2} + v^{2})} = \frac{1 x 10 x (50^{2} - 40^{2})}{(50^{2} + 40^{2})} N = \frac{90}{41} N$

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