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  • A bird is sitting on the top of 80 m high tree from a point on the ground the angle of elevation of the bird is 45 degree the bird flew away horizontally in such a way that it remained at a constan

A bird is sitting on the top of an 80 m high tree. From a point on the ground, the angle of elevation of the bird is $45^\circ$. The bird then flies away horizontally in such a way that it remains at a constant height from the ground. After $2$ seconds, the angle of elevation of the bird from the same point becomes $30^\circ$. Find the speed at which the bird is flying.

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Given the tree height is $h=80$ m.

When the angle of elevation is $45^\circ$, $ x_1=\dfrac{h}{\tan 45^\circ}=h $.

When the angle becomes $30^\circ$, $ x_2=\dfrac{h}{\tan 30^\circ}=h\sqrt{3} $.

Horizontal distance flown by the bird in $2$ s:

$ \Delta x = x_2 - x_1 = h(\sqrt{3}-1) $.

Substitute $h=80$: $ \Delta x = 80(\sqrt{3}-1) $ m.

Speed $v = \dfrac{\Delta x}{2} = \dfrac{80(\sqrt{3}-1)}{2} = 40(\sqrt{3}-1) $ m/s.

On putting the value of $\sqrt{3}=1.732$, we get, $ v \approx 40(1.7320508-1) \approx 29.28 $ m/s

Hence, the bird is flying at the speed of approximately $29.28$ m/s.

Posted by

Komal Miglani

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Posted by

Komal Miglani

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