JEE Main 2014 · 6 Apr · Shift 1 · Mathematics Q89

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MathematicsSingle correct+4 / -1Question 89 of 90

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

20\sqrt2

20\left(\sqrt3-1\right)

40\left(\sqrt2-1\right)

40\left(\sqrt3-\sqrt2\right)

Answer and worked solution

Official answer: B

The horizontal distance from O is $20\cot45^\circ=20$ m at first and $20\cot30^\circ=20\sqrt3$ m after 1 s. So the speed is $20\left(\sqrt3-1\right)$ m/s.

Topics:Heights and distances

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JEE Main 2014 · 6 Apr · Shift 1, Mathematics question 89