A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ftys along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole

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25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path. This can be obtained by considering this as a right angled triangle.

How fast is the tip of his shadow moving?

Let x be the length between man and the pole, y be the distance between the tip of the shadow and the pole.

Then y - x will be the length between the man and the tip of the shadow.

Since two triangles are similar, we can write

[tex]\frac{y-x}{y} =\frac{6}{15}[/tex]

⇒15(y-x) = 6y

15 y - 15 x = 6y

9y = 15x

y = 15/9 x

y = 5/3 x

Differentiate both sides

dy/dt = 5/3 dx/dt

dy/dt is the speed of the tip of the shadow, dx/dt is the speed of the man.

Given that dx/dt = 5 ft/s

Thus dy/dt = (5/3)×5 ft/s

dy/dt = 25/3 ft/s

Hence 25/3 ft/s is speed of the tip of his shadow moving when a man is 40 ft from the pole given that a street light is mounted at the top of a 15-ft-tall pole and the man is 6 ft tall who is walking away from the pole with a speed of 5 ft/s along a straight path.

Learn more about similar triangles here:

brainly.com/question/8691470

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