Question

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60 degree after 10 seconds acceleration is observed to be 30 degree find the speed of the aeroplane in kilometre per hour

Answer 1

Hey.

Height of the plane = 1 km

Initially elevation = 60°

after time t = 10 s

Final elevation = 60°

So, initially the distance OP1

= 1/tan60° = 1/√3 km

Similarly, final distance OP2'

= 1/tan30° = 1/(1/√3) = √3 km

{as in ∆OP'P2' tan30° = 1/OP2' }

so, displacement = OP2' - OP1

$\sqrt{3 } - \frac{1}{ \sqrt{3} } \\ \\ = \frac{(3 - 1)}{ \sqrt{3} } \\ \\ = \frac{ 2}{ \sqrt{3} } \: km$

Now t = 10 s = 10× 1/60×60 h = 10×1/3600 h

$so \: \: v \: = \: \frac{displacement}{time} \\ \\ so \: \: v \: = \: \frac{ \frac{2}{ \sqrt{3} } }{10 \times \frac{1}{3600} } \: \: \: km {h}^{ - 1} \\ \\ so \: \: v \: = \: \frac{720}{ \sqrt{3} } \: \: = \: 240 \sqrt{3} \: \: km {h}^{ - 1}$

Thanks.