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(III) A person stands at the base of a hill that is a straight incline making an angle $\phi$ with the horizontal (Fig. 48). For a given initial spced $v_{0},$ at what angle $\theta$ (to the horizontal)should objects be thrown so that the distance $d$ they land up the hill is as large as possible?

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$\theta=\frac{\phi}{2}+\frac{\pi}{4}$

Physics 101 Mechanics

Chapter 3

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So here we're gonna choose the origin to be at the bottom of the hill just where the incline starts. We consider that the equation of the equation of the line describing the hill weaken d notice why sub too. And this would equal X tangent of data, state of being, of course, the angle of incline of the hill. We consider the equations of the motion for the objects we can say Why? So are gonna be waifs of one equaling V Y initial t plus 1/2 a sub y t squared. And then we can say that X is gonna be equal to the ex initial t Ah, with the ex initial equaling the initial co sign of data. And of course, V. Why initial equaling the initial sign of data we can say we can solve for the horizontal equation, um, for the time of flight, and then insert that into the vertical motion equation so we can say, basically, t is gonna e be equal to X divided by V X initial, which would, of course, equal to X over the initial co sign of data. Now, we're gonna plug this tea into this equation. So this is great. So this tea term into the second equation here. So we can say that, uh, why sub one would be equal to of the initial sign of fada times X over the initial co sign of Fada Um minus well, say, uh G over two times X over the initial co sign of fada quantity squared. And at this point, recon say that why someone is simply gonna be equal to x tangent of beta minus, uh g X squared divided by two times the initial squared co sign squared of theta And at this point, we can set this equal to wise up to. So this would be ex tangent of theta equals extensions of data uh, minus again G X squared, divided by two be initial squared co sign Squared of Theta And then we can say that tangent of data minus tangent of fi Some angle is going to be equal to G X squared Divide by two initial squared co sign Squared of Theta Let's get a new workbook here and so for X So X will then be equal to Tangent of fate A minus tangent of five divided by G times to the initial squared co sign squared of theta and hear this intersection X coordinate is related to desired quantity D where x is equal ng d co sign of five So here we can set this equal to this equation now again and say that, um D co sign of fire is gonna equal tangent of data minus tangent of fi times to the initial squared co sign squared data divided by G and we're going to solve for D now. So d will then be equal to to be initial squared the bottom i g co sign of five multiplied by sine of data co sign of data a minus Tangent of fi co sign squared of theta. And in order to maximize distance, we need to set the derivative of this equation. Ah, with respect to theta to zero and then solve for theta. So we can say the derivative of D with respect to theta would be equal to two v initial squared divided by G Co sign of five times D over D theta times Sign of data co sign of Fada minus tangent of five co sign squared of data and we're going to set this equal. We rather this equals to the initial squared over G co sign of five on then take the derivative of this term here. This will be well, you have to use product roll so sign of fate at times Negative Sign of data plus co sign a theta Times co sign if ADA minus tangent of fi times two times co sign of data times Negative sign of Fada close parentheses And, uh, this is going to be equal. We can say this is gonna be equal to to be initial square to about a bi g co signed a fire once again keep the factor and most part by negative signs squared of theta plus co sign Squared of theta plus two tangent of five Co Sign of fate A sign if Ada and then this again is gonna be equal to To the initial squared Divided by Geico Sign if I Times Co Sign of tooth data plus sign of tooth Ada Times Tangent of five Set this equal to zero and solve for theta so we can see that co sign of tooth Ada plus sign of tooth Ada Times Tangent of five will equal zero their fourth date. I will equal 1/2 times arc tanne of negative one over Tangent of fire. So this would be our answer. That is the end of the solution. Thank you for watching.

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