A car with a vertical windsheld moves in a rain storm at a speed of $40 k m / h r$ . The rain drops fall vertically with constant speed of $20 m / s$ . The angle at which rain drops strike the windsheild is

{tan}^{- 1} \frac{5}{9}

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{tan}^{- 1} \frac{9}{5}

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{tan}^{- 1} \frac{3}{2}

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{tan}^{- 1} \frac{2}{3}

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Solution

The correct option is A ${tan}^{- 1} \frac{5}{9}$
The horizontal component of falling rain is $40 k m / h r$
Convert $k m / h r$ into $m / s$
$40 k m / h r = 40 \times \frac{5}{18} m / s = \frac{200}{18} m / s$
The vertical component of falling rain is $20 m / s$
Now,
$Q = {tan}^{- 1} (\frac{H o r i z o n t a l}{V e r t i c a l})$
$Q = {tan}^{- 1} (\frac{200 / 18}{20})$
$Q = {tan}^{- 1} (\frac{5}{9})$

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A car with a vertical windsheld moves in a rain storm at a speed of 40km/hr. The rain drops fall vertically with constant speed of 20m/s. The angle at which rain drops strike the windsheild is

The correct option is A tan−159The horizontal component of falling rain is 40km/hrConvert km/hr into m/s 40km/hr=40×518m/s=20018m/sThe vertical component of falling rain is 20m/sNow,Q=tan−1(HorizontalVertical)Q=tan−1(200/1820)Q=tan−1(59)

A car with a vertical windsheld moves in a rain storm at a speed of $40 k m / h r$ . The rain drops fall vertically with constant speed of $20 m / s$ . The angle at which rain drops strike the windsheild is