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2 years ago

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SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the

building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building

1 answer:

Firlakuza [10]2 years ago

5 0

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

H = 2 m × 12 m/(12 m - 4 m)

H = 24 m²/8 m

H = 3 m

Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

dL/dt = h(dD/dt)/H

Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)

Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

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It is given that energy in first exited state (n=2) , $E_2$ = -3.4 ev

Also , it is given than energy in ground state (n=1) , $E_1$ = -13.6 ev.

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$E_{joule}=\dfrac{h\times c}{\lambda}$

where , $h=6.626\times 10^{-34}\ joule-second$

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$10.2 \times 1.6\times 10^{-19}=\dfrac{6.626\times10^{-34} \times 3 \times 10^8}{\lambda}$

$\lambda=\dfrac{{6.626\times10^{-34} \times 3 \times 10^8}}{10.2 \times 1.6\times 10^{-19}}$

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Therefore, $frequency=\dfrac{c}{\lambda}= \dfrac{3\times 10^8}{1.21\times 10^-7} \ s^{-1}=2.479\times 10^{15} \ s^{-1}$

Hence, this is the required solution.

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