A bird is perched on the top of a tree $ 20 \mathrm{~m} $ high and its elevation from a point on the ground is $ 45^{\circ} $. It flies off horizontally straight away from the observer and in one second the elevation of the bird is reduced to $ 30^{\circ} $. The speed of the bird is -

Question

A bird is perched on the top of a tree $ 20 \mathrm{~m} $ high and its elevation from a point on the ground is $ 45^{\circ} $. It flies off horizontally straight away from the observer and in one second the elevation of the bird is reduced to $ 30^{\circ} $. The speed of the bird is -, $ 14.64 \mathrm{~m} / \mathrm{s} $, 17. $ 71 \mathrm{~m} / \mathrm{s} $, $ 12 \mathrm{~m} / \mathrm{s} $, None of these

MARKS App · Accepted Answer

Let the bird be perched at B , the top of the tree B D and O be the observer. then, ∠ B O D = 45 ° and B D = 20 m . Now, the bird flying horizontally reaches M in 1 s . Then, ∠ M O N = 30 ° , where, M N ⊥ O N Now, B D = M N = 20 m From triangle B O D . tan 4 5 ° = B D O D = 2 0 O D ⇒ O D = 20 m Now, from Δ M O N tan 3 0 ° = M N O N = 2 0 2 0 + D N ⇒ 1 3 = 2 0 2 0 + D N ⇒ 2 0 + D N = 2 0 3 ⇒ D N = 2 0 3 - 1 = 20 x 0 . 732 = 14 . 64 m Now, Speed = distance time = B M 1 = D N 1 = 14 . 64 m / s

Search any question & find its solution

Looking for more such questions to practice?