Answer to Question #105168 in Trigonometry for Jannat friend

Question

A street light is at the top of a 11 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 45 ft from the base of the pole?

Answer 1

$\frac{OD}{AD}=\frac{H}{h}$ or $\frac{OD}{OD-v_1t}=\frac{H}{h}$ hence $OD=\frac{v_1Ht}{H-h}$ , $OD'=\frac{v_1H(t+\Delta t)}{H-h}$ then

$OD'-OD=\frac{Hv_1\Delta t}{H-h}$ $=v_2\Delta t$ hence $v_2=\frac{Hv_1}{H-h}\approx15.4\times0.305\frac{m}{s}\approx4.7\frac{m}{s}$ where $v_2$ is the speed of the point of shadow, $H=11ft, h=6ft, v_1=7\frac{ft}{s}$