Rain is falling vertically $4 m s^{- 1}$ . A man is moving due east with $3 m s^{- 1}$ . The direction in which he shall hold the umbrella with the vertical is

53^{\circ}

east of vertical

No worries! We‘ve got your back. Try BYJU‘S free classes today!

37^{\circ}

east of vertical

No worries! We‘ve got your back. Try BYJU‘S free classes today!

53^{\circ}

west of vertical

No worries! We‘ve got your back. Try BYJU‘S free classes today!

37^{\circ}

west of vertical

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B $37^{\circ}$ west of vertical
$r =$ rain $, g =$ ground $& m =$ man
We have, $_{r m} =_{r g} -_{m g}$
Resultant $_{r m}$ makes an angle of $θ = {tan}^{- 1} \frac{3}{4} o r θ = 37^{o}$
Man should hold his umbrella at $θ = 37^{o}$ west of vertical

Suggest Corrections

Similar questions

Q. Rain is falling vertically with

3 {ms}^{- 1}

and a man is moving due North with

4 {ms}^{- 1}

. In which direction he should hold the umbrella to protect himself from rains?

Rain is falling vertically with a speed of $12 m s^{- 1}$ .A cyclist is moving east to west with a speed of $12 \sqrt{3} m s^{- 1}$ .In order to protect himself from rain at what angle he should hold his umbrella w.r.t vertical.

Q. Rain is falling vertically with

3 m s^{- 1}

and a man is moving due North with

4 m s^{- 1}

. In which direction he should hold the umbrella to protect himself from rains?

Q. Rain is falling vertically with a velocity of

15 m s^{- 1}

. A person rides a bicycle with a speed of

15 m s^{- 1}

in the west to east direction. What is the direction (angle with vertical) in which he should hold his umbrella to save himself from the rain?

Q. A man rides a bicycle with a speed of

17.32

m s^{- 1}

East-West direction. If the rain falls vertically with a speed of

10

m s^{- 1}

, then the direction in which he must hold his umbrella is:

Join BYJU'S Learning Program

Rain is falling vertically 4 ms−1. A man is moving due east with 3 ms−1 . The direction in which he shall hold the umbrella with the vertical is

The correct option is B 37∘ west of verticalr=rain,g= ground &m=manWe have, →vrm=→vrg−→vmgResultant →vrm makes an angle of θ=tan−134orθ=37oMan should hold his umbrella at θ=37o west of vertical

Rain is falling vertically $4 m s^{- 1}$ . A man is moving due east with $3 m s^{- 1}$ . The direction in which he shall hold the umbrella with the vertical is

The correct option is B $37^{\circ}$ west of vertical
$r =$ rain $, g =$ ground $& m =$ man
We have, $_{r m} =_{r g} -_{m g}$
Resultant $_{r m}$ makes an angle of $θ = {tan}^{- 1} \frac{3}{4} o r θ = 37^{o}$
Man should hold his umbrella at $θ = 37^{o}$ west of vertical