Question

The angle of elevation of an aeroplane from a point on the ground is 45 degree. After 15 seconds of flight the angle changes to 30 degree.If the plane is flying at a constant height of 2500m,find the speed of the plane.

Answer 1

therefore, the speed of the aeroplane= 122 m/s

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Hope it helps...

Answer 2

Answer:

$Speed\:of\:the\:plane\\=\frac{500(\sqrt{3}-1)}{3}\:m/s$

Step-by-step explanation:

In right Triangle AMP,

$tan45\degree = \frac{AM}{PM}$

$\implies 1=\frac{2500}{PM}$

$\implies PM = 2500\:m$

$In \: right \: triangle \:BNP,\\tan30\degree = \frac{BN}{PN}$

$\implies \frac{1}{\sqrt{3}}=\frac{2500}{PN}$

$\implies PN = 2500\sqrt{3}$

Therefore,

$AB = MN=PN-PM\\=2500\sqrt{3}-2500\\=2500(\sqrt{3}-1)\:m$

Now ,

$Speed\:of\:the\:plane =\frac{distance}{time}$

$=\frac{AB}{15}\\=\frac{2500(\sqrt{3}-1)\:m}{15}$

$=\frac{500(\sqrt{3}-1)}{3}\:m/s$

Therefore,

$Speed\:of\:the\:plane\\=\frac{500(\sqrt{3}-1)}{3}\:m/s$

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