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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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Rain is falling vertically with a speed of $20\;ms^{-1}$. A person is running in rain (from east) with a velocity of $5 ms^{-1}$ and a wind is blowing with speed of $15ms ^{-1}$ (from west).The angle with the vertical at which the person should hold his umbrella is

\[(a)\;\tan ^{-1}\bigg(\frac{1}{2}\bigg)\quad (b)\;\tan ^{-1}\bigg(\frac{1}{3}\bigg) \quad (c)\;\tan ^{-1}\bigg(\frac{4}{5}\bigg) \quad (d)\;\tan ^{-1}(3)\]

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Velocity of rain $=v_r=20 j$
Velocity of man $=v_m=5 i$
Velocity of wind $=15 i$
Velocity of rain with respect to man when wind is blowing is
$\quad=15 i-5i+20j$
$\quad=20j+10 i$
$\tan \theta=\large\frac{20}{10}$
$\qquad=2$ from the horizontal.
Hence the person should hold is umbrella at $ \alpha$ (with vertical)
$\quad=\tan ^{-1}\bigg(\large\frac{1}{2}\bigg)$
Hence a is the correct answer
answered Jun 28, 2013 by meena.p
edited May 24, 2014 by lmohan717

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