First slide

Projection Under uniform Acceleration

Question

Rain is falling vertically. A man walking horizontally holds his umbrella at an angle of $45^{o}$ with vertical. If the man doubles his velocity, at what angle with vertical should be hold his umbrella?

Moderate

A

$\tan^{- 1} (\sqrt{5})$
B

$\tan^{- 1} (\frac{1}{\sqrt{5}})$
C

$\tan^{- 1} (\frac{1}{2})$
D

$\tan^{- 1} (2)$

Solution

${\vec{V}}_{R} = {\vec{V}}_{m} + {\vec{V}}_{\frac{R}{m}}$
$\tan α = \frac{V_{m}}{V_{R}}$
and $\tan α^{1} = \frac{V_{m}}{V_{R}} = 2 \frac{V_{m}}{V_{R}} = 2 \tan α = 2 \tan 45^{o}$
$\Rightarrow α^{1} = \tan^{- 1} (2)$

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