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Helo,

I've been working out this exercise, but my solution and the text's aren't the same.

We have a spotlight on the floor located at a distance of 30.5 m from a wall of a building. There is a person 1.83 m tall between the spotlight and the wall moving away from the spotlight at a constant velocity of 1.83 m/s. Calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight.

So, lets name

[itex]x_{1}=9.14 m[/itex]

[itex]x_{2}=30.5 m[/itex]

[itex]y_{1}=[/itex]the height of the shadow when the person is on [itex]x_{1}[/itex]

[itex]y_{2}=[/itex]the height of the shadow when the person is on [itex]x_{2}[/itex]

[itex]h=1.83 m[/itex] (height of the person)

[itex]v=1.83 m/s[/itex] (velocity)

1) I find the height of the shadow projected on the wall when the body is on x

[itex]\alpha=arctan(\frac{h}{x_{1}})[/itex]

And now I find [itex]y_{1}[/itex]

[itex]y_{1}=tan \alpha·x_{2}=6.11 m[/itex]

Then the height of the shadow at [itex]x_{2}[/itex] will be [itex]y_{2}=1.83 m[/itex]. I think this is obvious.

Afterwards, I want to find the time that took place moving from [itex]x_{1}[/itex] to [itex]x_{2}[/itex] because this will be the same time from [itex]y_{1}[/itex] to [itex]y_{2}[/itex]

[itex]\Delta t=\frac{\Delta x}{v}=11.67 s[/itex]

Then, to conclude

[itex]v_{shadow}=\frac{\Delta y}{\Delta t}=-0.366 m/s[/itex]

The solution of the book is

[itex]v_{shadow}=-1.22 m/s[/itex]

Anyone can help me, please?

I've been working out this exercise, but my solution and the text's aren't the same.

**Homework Statement**We have a spotlight on the floor located at a distance of 30.5 m from a wall of a building. There is a person 1.83 m tall between the spotlight and the wall moving away from the spotlight at a constant velocity of 1.83 m/s. Calculate the velocity of the shadow variation when the person is at 9.14 m from the spotlight.

So, lets name

[itex]x_{1}=9.14 m[/itex]

[itex]x_{2}=30.5 m[/itex]

[itex]y_{1}=[/itex]the height of the shadow when the person is on [itex]x_{1}[/itex]

[itex]y_{2}=[/itex]the height of the shadow when the person is on [itex]x_{2}[/itex]

[itex]h=1.83 m[/itex] (height of the person)

[itex]v=1.83 m/s[/itex] (velocity)

**The attempt to a solution**1) I find the height of the shadow projected on the wall when the body is on x

_{1}. To do this, I consider the spotlight and the person make a right triangle, with angle [itex]\alpha[/itex] which is the same of the triangle made with the height of the shadow y_{1}and the distance between the spotlight and the wall.[itex]\alpha=arctan(\frac{h}{x_{1}})[/itex]

And now I find [itex]y_{1}[/itex]

[itex]y_{1}=tan \alpha·x_{2}=6.11 m[/itex]

Then the height of the shadow at [itex]x_{2}[/itex] will be [itex]y_{2}=1.83 m[/itex]. I think this is obvious.

Afterwards, I want to find the time that took place moving from [itex]x_{1}[/itex] to [itex]x_{2}[/itex] because this will be the same time from [itex]y_{1}[/itex] to [itex]y_{2}[/itex]

[itex]\Delta t=\frac{\Delta x}{v}=11.67 s[/itex]

Then, to conclude

[itex]v_{shadow}=\frac{\Delta y}{\Delta t}=-0.366 m/s[/itex]

The solution of the book is

[itex]v_{shadow}=-1.22 m/s[/itex]

Anyone can help me, please?

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