A bird is sitting on the top of a vertical po | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the initial position of the bird be $P$ at the top of the pole $PQ = 20 \ m$. The angle of elevation from point $O$ is $\angle POQ = 45^o$.In

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let the initial position of the bird be $P$ at the top of the pole $PQ = 20 \ m$. The angle of elevation from point $O$ is $\angle POQ = 45^o$.In $\Delta POQ$,$	an 45^o = \frac{PQ}{OQ}$ $\Rightarrow 1 = \frac{20}{OQ}$ $\Rightarrow OQ = 20 \ m$.After one second,the bird reaches position $P'$ such that $P'Q' = 20 \ m$ (since it flies horizontally). The angle of elevation from $O$ is $\angle P'OQ' = 30^o$.In $\Delta P'OQ'$,$	an 30^o = \frac{P'Q'}{OQ'}$ $\Rightarrow \frac{1}{\sqrt{3}} = \frac{20}{OQ'}$ $\Rightarrow OQ' = 20\sqrt{3} \ m$.The distance covered by the bird in one second is $QQ' = OQ' - OQ = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1) \ m$.Since the time taken is $1 \ s$,the speed of the bird is $\frac{20(\sqrt{3} - 1)}{1} = 20(\sqrt{3} - 1) \ m/s$.

Similar Questions

Two vertical poles $AB = 15 \ m$ and $CD = 10 \ m$ are standing apart on a horizontal ground with points $A$ and $C$ on the ground. If $P$ is the point of intersection of $BC$ and $AD$,then the height of $P$ (in $m$) above the line $AC$ is:

An aeroplane flying with uniform speed horizontally $1 \ km$ above the ground is observed at an elevation of $60^{\circ}$. After $10 \ s$,if the elevation is observed to be $30^{\circ}$,then the speed of the plane (in $km/h$) is

From a point on the level ground,the angle of elevation of the top of a pole is $30^{\circ}$. On moving $20 \ m$ nearer to the pole,the angle of elevation becomes $45^{\circ}$. The height of the pole (in metres) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module