JEE MAIN - Physics (2019 - 12th January Morning Slot - No. 21)
A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is :
$${{\sqrt 3 } \over 2}$$v
$${{2v} \over {\sqrt 3 }}$$
v
$${v \over 2}$$
Explanation
_12th_January_Morning_Slot_en_21_1.png)
AB = VP $$ \times $$ t
BC = Vt
cos60o = $${{AB} \over {BC}}$$
$${1 \over 2} = {{{V_P} \times t} \over {Vt}}$$
VP = $${V \over 2}$$
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