A balloon is descending at a rate of 4 , m/mi | vedclass

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(B) Let the distance of the balloon from the observer be $d$. Let the initial height be $h_1$ and the final height be $h_2$.From the geometry,$h_1 = d

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$20(3 + \sqrt{3}) \, m$

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(B) Let the distance of the balloon from the observer be $d$. Let the initial height be $h_1$ and the final height be $h_2$.From the geometry,$h_1 = d 	an 45^\circ = d$ and $h_2 = d 	an 30^\circ = \frac{d}{\sqrt{3}}$.The distance descended in $10 \, minutes$ is $h_1 - h_2 = d - \frac{d}{\sqrt{3}} = d \left( \frac{\sqrt{3} - 1}{\sqrt{3}} ight)$.Given the rate of descent is $4 \, m/min$,the distance descended in $10 \, minutes$ is $4 	imes 10 = 40 \, m$.Equating the two: $d \left( \frac{\sqrt{3} - 1}{\sqrt{3}} ight) = 40$.$d = \frac{40 \sqrt{3}}{\sqrt{3} - 1} = \frac{40 \sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{40(3 + \sqrt{3})}{2} = 20(3 + \sqrt{3}) \, m$.

$A$ balloon is descending at a rate of $4 \, m/min$. Its angle of elevation from a point on the ground is $45^\circ$,which reduces to $30^\circ$ after $10 \, minutes$. At what distance from the observer will the balloon touch the ground?

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