Question

One evening, Ratan was in a park. Children were playing cricket. Birds were singing on a nearby tree of height 80 m. He observed a bird on the tree at an angle of elevation of 45°. When a sixer was hit, a ball flew through the tree, frightening the bird to fly away. In 2 seconds, he observed the bird flying at the same height at an angle of elevation of 60°.

Based on the above information, answer the following questions.

1. At what distance from the foot of the tree was Ratan observing the bird sitting on the tree?
2. How far did the bird fly in the mentioned time?
3. After hitting the tree, how far did the ball travel in the sky when Ratan saw it?
4. What is the speed of the bird in metre/minute if it has flown \(20(\sqrt{3}+1) \mathrm{~m}\)?

Answer 1

Let the tree be CB of height 80 m, with B as the foot. Ratan is at point A on the ground.

i. Initial distance from the foot of the tree

In right \(\triangle CBA\), \(\angle CAB = 45^\circ\):

\[\tan 45^\circ = \frac{CB}{BA}\]

\[\Rightarrow 1 = \frac{80}{BA}\]

\[\Rightarrow BA = 80 \text{ m}\]

Ratan was 80 m from the foot of the tree.

ii. Distance the bird flew

Let the bird fly from C to E same height 80 m, with \(\angle EAD = 30^\circ\) where D is vertically below E. Then ED = 80 m.

In right \(\triangle EDA\): \[\cot 30^\circ = \frac{AD}{ED}\]

\[\Rightarrow \sqrt{3} = \frac{AD}{80}\]

\[\Rightarrow AD = 80\sqrt{3} \text{ m}\]

Distance flown by bird = \(CD = AD - AC\)

\(= 80\sqrt{3} - 80\)

\(= 80(\sqrt{3} - 1)\) m

iii. Distance the ball travelled when Ratan saw it at 60° elevation

Let the ball be at G same height 80 m with \(\angle GAF = 60^\circ\) where F is vertically below G. Then GF = 80 m.

In right \(\triangle GFA\): \[\cot 60^\circ = \frac{AF}{GF}\]

\[\Rightarrow \frac{1}{\sqrt{3}} = \frac{AF}{80}\]

\[\Rightarrow AF = \frac{80}{\sqrt{3}} \text{ m}\]

Distance travelled by ball = \(CF = AC - AF\)

\(= 80 - \frac{80}{\sqrt{3}}\)

\(= 80\left(1 - \frac{1}{\sqrt{3}}\right)\) m.

iv. Speed of the bird if it flew \(20(\sqrt{3} + 1)\) m in 2 seconds

\[\text{Speed} = \frac{20(\sqrt{3} + 1) \text{ m}}{2 \text{ s}}\]

\[= 10(\sqrt{3} + 1) \text{ m/s}\]

Convert to m/min: \[10(\sqrt{3} + 1) \times 60 = 600(\sqrt{3} + 1) \text{ m/min}\]

Speed = \(600(\sqrt{3} + 1)\) m/min.