A parachutist is descending vertically and makes angles of elevation of $ 45^{\circ} $ and $ 60^{\circ} $ at two observing points $ 100 \mathrm{~m} $ apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.

Given:

A parachutist is descending vertically and makes angles of elevation of $ 45^{\circ} $ and $ 60^{\circ} $ at two observing points $ 100 \mathrm{~m} $ apart from each other on the left side of himself.

To do:

We have to find the maximum height from which he falls and the distance of the point where he falls on the ground from the just observation point.

Solution:

Let $AB$ be the height of the parachutist above the ground and $C, D$ be the points of observation which make angles of elevation of $ 45^{\circ} $ and $ 60^{\circ} $ respectively.

From the figure,

$\mathrm{CD}=100 \mathrm{~m}, \angle \mathrm{ACB}=45^{\circ}, \angle \mathrm{ADB}=60^{\circ}$

Let the height of the parachutist above the ground be $\mathrm{AB}=h \mathrm{~m}$ and the distance between the point $D$ and the the point where he falls on the ground be $\mathrm{DB}=x \mathrm{~m}$.

This implies,

$\mathrm{CB}=x+100 \mathrm{~m}$

We know that,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{BC}$

$\Rightarrow \tan 45^{\circ}=\frac{h}{x+100}$

$\Rightarrow 1=\frac{h}{x+100}$

$\Rightarrow h=1(x+100) \mathrm{~m}$

$\Rightarrow x=h-100 \mathrm{~m}$...........(i)

Similarly,

$\tan \theta=\frac{\text { Opposite }}{\text { Adjacent }}$

$=\frac{\text { AB }}{DB}$

$\Rightarrow \tan 60^{\circ}=\frac{h}{x}$

$\Rightarrow \sqrt3=\frac{h}{x}$

$\Rightarrow x(\sqrt3)=h \mathrm{~m}$

$\Rightarrow (h-100)\sqrt3=h \mathrm{~m}$ [From (i)]

$\Rightarrow \sqrt3h-100\sqrt3=h \mathrm{~m}$

$\Rightarrow h(\sqrt3-1)=100\sqrt3 \mathrm{~m}$

$\Rightarrow h=\frac{100\times1.732}{1.732-1} \mathrm{~m}$

$\Rightarrow h=\frac{173.2}{0.732}=236.6 \mathrm{~m}$

$\Rightarrow x=236.6-100=136.6 \mathrm{~m}$

Therefore, the maximum height from which he falls is $236.6 \mathrm{~m}$ and the distance of the point where he falls on the ground from the just observation point is $136.6 \mathrm{~m}$.

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Updated on: 10-Oct-2022

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