If a man of height 1.8 mt . is walking away from the foot of a light pole of height 6 mt . with a speed of 7 km per hour on a straight horizontal road opposite to the pole, then the rate of change of the length of his shadow is (in kmph)

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Aicharya · Accepted Answer

Let OA be the light pole, GF be the man standing at G after time t . In $\triangle \mathrm{AEF}, \tan \theta=\frac{4.2}{x}$

In $\triangle \mathrm{AOB}, \tan \theta=\frac{6}{x+y}$ $\Rightarrow \frac{4.2}{x}=\frac{6}{x+y} \Rightarrow 1.8 x=4.2 y$
Differentiating w.r.t. t $; 1.8 \frac{d x}{d t}=4.2 \frac{d y}{d t}$ $\frac{1.8 \times 7}{4.2}=\frac{d y}{d t} \Rightarrow \frac{d y}{d t}=3 \mathrm{kmph}$

If a man of height 1.8 mt . is walking away from the foot of a light pole of height 6 mt . with a speed of 7…

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