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Lyrx [107]
3 years ago
9

A 7 ft tall person is walking away from a 20 ft tall lamppost at a rate of 5 ft/sec. Assume the scenario can be modeled with rig

ht triangles. At what rate is the length of the person's shadow changing when the person is 16 ft from the lamppost?
Physics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

Answer:

The rate of change of the shadow length of a person is 2.692 ft/s

Solution:

As per the question:

Height of a person, H = 20 ft

Height of a person, h = 7 ft

Rate = 5 ft/s

Now,

From Fig.1:

b = person's distance from the lamp post

a = shadow length

Also, from the similarity of the triangles, we can write:

\frac{a + b}{20} = \frac{a}{7}

a = \frac{7}{13}b

Differentiating the above eqn w.r.t t:

\frac{da}{dt} = \frac{7}{13}.\frac{db}{dt}

Now, we know that:

Rate = \frac{db}{dt} = 5\ ft/s

Thus

\frac{da}{dt} = \frac{7}{13}.\times 5 = 2.692\ ft/s

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Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

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Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

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                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

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Answer:

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