Answer to Question #105168 in Trigonometry for Jannat friend

Question #105168
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?
1
Expert's answer
2020-03-16T11:48:25-0400

"\\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}"



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS