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Introductory Physics Homework Help
Find the velocity of the shadow variation
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[QUOTE="adriaat, post: 4514777, member: 443535"] Helo, I've been working out this exercise, but my solution and the text's aren't the same. [b]Homework Statement [/b] 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, let's 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) [b]The attempt to a solution[/b] 1) I find the height of the shadow projected on the wall when the body is on x[SUB]1[/SUB]. 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[SUB]1[/SUB] 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? [/QUOTE]
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Introductory Physics Homework Help
Find the velocity of the shadow variation
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