Question

The angles of elevation of a balloon from the observer’s eye at two instances are 45º and 60º. If the height of balloon in both the instants is 10 m and it takes $(\sqrt{3}-1)$ seconds, then the approximate speed of the balloon is $(\text{Use}\sqrt{3}=1.73)$

• A. 4.8 m/s
• B. 5 m/s
• C. 6.4 m/s
• D. 5.8 m/s

Answer 1

The correct option is D 5.8 m/s

Let A and E be the positions of the balloon which makes an elevation of 60° and 45° respectively from the observer’s eye at point B.
$In\Delta ACB,$
$\tan {60}^{\circ }=\frac{AC}{BC}$
$\Rightarrow \sqrt{3}=\frac{10}{BC}$
$\Rightarrow BC=\frac{10}{\sqrt{3}}m$
$Similarly,in\Delta EDB,$
$\tan {45}^{\circ }=\frac{ED}{BD}$
$\Rightarrow 1=\frac{10}{BD}$ $(\because ED=AC=10m)$
$\Rightarrow BD=10m$
$Thus,CD=BD-BC=(10-\frac{10}{\sqrt{3}})m$
Now, time taken by balloon to reach from C to D is $(\sqrt{3}-1)$ seconds.
$\therefore Speedoftheballoon=\frac{Dis\tan ce}{Time}$
$=\frac{10-\frac{10}{\sqrt{3}}}{\sqrt{3}-1}$
$=\frac{\frac{10(\sqrt{3}-1)}{\sqrt{3}}}{\sqrt{3}-1}$
$=\frac{10}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{10\sqrt{3}}{\sqrt{3}}$
$=\frac{10\times 1.73}{3}$
= 5.8 m/s
Therefore, the speed of the balloon is 5.8 m/s.
Hence, the correct answer is option (d).