Discussion Forum



1)

 A bird is sitting on the top of vertical pole 20m high and its elevation from a point O on the ground 45° . It flies off horizontally straight away from the point O. After 1s,  the elevatiion  of the bird from O is reduced to 30° . Then the speed (in m/s)  of the bird is 


A) $40(\sqrt{3}-1)$

B) $20\sqrt{2}$

C) $40(\sqrt{3}-\sqrt{2})$

D) $20(\sqrt{3}-1)$

Answer:

Option D

Explanation:

In  $\triangle OA_{1}B_{1},$

    $\tan 45^{0}=\frac{A_{1}B_{1}}{OB_{1}}\Rightarrow \frac{20}{OB_{1}}=1$

  $\Rightarrow$      $ OB_{1}=20$

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$\triangle OA_{2}B_{2},$

   $\tan 30^{0}=\frac{20}{OB_{2}}$

 $\Rightarrow$           $OB_{2}=20\sqrt{3}$

$\Rightarrow$                     $B_{1}B_{2}+OB_{1}=20\sqrt{3}$

 $\Rightarrow$                    $B_{1}B_{2}=20\sqrt{3}-20$            

 $\Rightarrow$               $B_{1}B_{2}=20(\sqrt{3}-1)m$

 Now, speed = $\frac{Distance}{Time}=\frac{20(\sqrt{3}-1)}{1}$

                              =   $20(\sqrt{3}-1)$  m/s