Question

raindrops falling with actual velocity of 5m/s on a windy day appeared to fall vertically down on a man moving eastwards with a speed of 3ms-¹ the man reversed his direction in the same speed thr rain drops will fall on him with a speed of​

Answer 1

Given:

Raindrops falling with actual velocity of 5m/s on a windy day appeared to fall vertically down on a man moving eastwards with a speed of 3ms-¹ the man reversed his direction in the same speed.

To find:

New velocity of rain with respect to man?

Calculation:

When man was moving east:

• Refer to 1st diagram:

$\therefore \: \sin( \theta) = \dfrac{3}{5}$

$\implies \: \theta \approx {37}^{ \circ}$

Now , when man is moving west:

• Refer to 2nd diagram.

• Angle between v_(m) and v_(r) becomes 90+37 = 127°.

Let new velocity of rain w.r.t man be v_(rm):

$\therefore \: v_{rm} = \sqrt{ {(v_{m})}^{2} + {(v_{r})}^{2} + 2(v_{m})(v_{r}) \cos( {180}^{ \circ} - {127}^{ \circ} ) }$

$\implies \: v_{rm} = \sqrt{ {(v_{m})}^{2} + {(v_{r})}^{2} + 2(v_{m})(v_{r}) \cos( {53}^{ \circ} ) }$

$\implies \: v_{rm} = \sqrt{ {(3)}^{2} + {(5)}^{2} + 2(3)(5) \cos( {53}^{ \circ} ) }$

$\implies \: v_{rm} = \sqrt{ 34 + 30\cos( {53}^{ \circ} ) }$

$\implies \: v_{rm} = \sqrt{ 34 + (30 \times \dfrac{3}{5} ) }$

$\implies \: v_{rm} = \sqrt{ 34 +18}$

$\implies \: v_{rm} = \sqrt{52}$

$\implies \: v_{rm} \approx \: 7.21 \: m {s}^{ - 1}$

So, new velocity of rain w.r.t man is 7.21 m/s.