An aeroplane at an altitude of 1 text{ km} is | vedclass

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(A) Let $x$ be the horizontal distance and $z$ be the straight-line distance from the observer to the aeroplane.Given altitude $h = 1 	ext{ km}$. By

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$360$

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(A) Let $x$ be the horizontal distance and $z$ be the straight-line distance from the observer to the aeroplane.Given altitude $h = 1 	ext{ km}$. By Pythagoras theorem, $z^2 = x^2 + h^2$, so $z^2 = x^2 + 1^2$.Differentiating with respect to time $t$: $2z \frac{dz}{dt} = 2x \frac{dx}{dt} \implies z \frac{dz}{dt} = x \frac{dx}{dt}$.Given $z = 1250 	ext{ m} = 1.25 	ext{ km}$ and $\frac{dx}{dt} = 600 	ext{ km/hr}$.Calculate $x$ when $z = 1.25 	ext{ km}$: $x = \sqrt{z^2 - 1^2} = \sqrt{1.25^2 - 1^2} = \sqrt{1.5625 - 1} = \sqrt{0.5625} = 0.75 	ext{ km}$.Substitute values into the derivative equation: $1.25 \frac{dz}{dt} = 0.75 	imes 600$.$\frac{dz}{dt} = \frac{450}{1.25} = 360 	ext{ km/hr}$.

An aeroplane at an altitude of $1 \text{ km}$ is flying horizontally at $600 \text{ km/hr}$, passes directly over an observer. The rate at which it is approaching the observer when it is $1250 \text{ meters}$ away from him is........ (in $\text{ km/hr}$)

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