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Q. 4.25 An aircraft is flying at a height of  3400\; m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0\; s apart is 30^{\circ}, what is the speed of the aircraft?

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The given situation is shown in the figure:-

                              Motion in a plane ,     20230

For finding the speed of aircraft we just need to find the distance AC as we are given t = 10 sec.

Consider \Delta ABD,

                                         \tan 15^{\circ}\ =\ \frac{AB}{BD}

                                           AB\ =\ \ BD\ \times \tan 15^{\circ}

or                                       AC\ =\ 2AB\ =\ \ 2BD\ \times \tan 15^{\circ}

or                                                  =\ \ 2\times 3400\times \tan 15^{\circ}

or                                                  =\ 1822.4\ m

Thus, speed of aircraft  :

                                            =\ \frac{1822.4}{10}\ =\ 182.24\ m/s

Posted by

Devendra Khairwa

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