Question:
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 $10\,s$ apart is $30^\circ$ then the speed of the aircraft is
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 $10\,s$ apart is $30^\circ$ then the speed of the aircraft is
Updated On: Mar 4, 2024
- $ 19.63\,m{{s}^{-1}} $
- $ 196.3\,m{{s}^{-1}} $
- $ 108\,m{{s}^{-1}} $
- $ 1963\,m{{s}^{-1}} $
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The Correct Option is B
Solution and Explanation
$O$ is the observation point at the ground. $A$ and $B$ are the positions of aircraft for which $? AOB=30^?$.
Time taken by aircraft from $A$ to $B$ is $10\,s$.
In $\Delta AOB,$
$tan\,30^{?}$
$=\frac{AB}{3400}$
$AB=3400\,tan\,30^{?}$
$=\frac{3400}{\sqrt{3}}m$
$\therefore$ Speed of aircraft, $v=\frac{AB}{10}$
$=\frac{3400}{10\sqrt{3}}$
$=196.3\,ms^{-1}$
Time taken by aircraft from $A$ to $B$ is $10\,s$.
In $\Delta AOB,$
$tan\,30^{?}$
$=\frac{AB}{3400}$
$AB=3400\,tan\,30^{?}$
$=\frac{3400}{\sqrt{3}}m$
$\therefore$ Speed of aircraft, $v=\frac{AB}{10}$
$=\frac{3400}{10\sqrt{3}}$
$=196.3\,ms^{-1}$
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Concepts Used:
Motion in a Plane
It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions.
Equations of Plane Motion
The equations of motion in a straight line are:
v=u+at
s=ut+½ at2
v2-u2=2as
Where,
- v = final velocity of the particle
- u = initial velocity of the particle
- s = displacement of the particle
- a = acceleration of the particle
- t = the time interval in which the particle is in consideration