Answer:
Let B be the initial position of bird sitting on top of tree of length 80 m. After 2 sec, the position of bird becomes C. Let the distance travel by bird from B to C is x m. Now, in \[\Delta \text{ }ABO\] \[\tan \,45{}^\circ =\frac{AB}{AO}=\frac{80}{y}\] \[y=80\,m\] ?(i) And, in \[\Delta \,DCO\] \[\tan \,30{}^\circ =\frac{CD}{DO}=\frac{80}{x+y}\] \[\frac{1}{\sqrt{3}}=\frac{80}{x+80}\] [Using eq. (i)] \[x+80=80\sqrt{3}\] \[x=80\left( \sqrt{3}-1 \right)=80\times 0.732\] \[\therefore x=58.56\,m\] Hence, speed of flying of the bird \[=\frac{58.56}{2}\] \[\left( Speed=\frac{Dis\tan ce}{Time} \right)\] \[=29.28\,m/s\]
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