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Carnegie Mellon University

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Problem 46 Hard Difficulty

The $x$ - and $y$ -coordinates of a projectile launched from the origin are $x=v_{0 x} t$ and $y=v_{0 y} t-\frac{1}{2} g t^{2}$ . Solve the first of these equations for time $t$ and substitute into the second to show that the path of a projectile is a parabola with the form $y=a x+b x^{2},$ where $a$ and $b$ are constants.

Answer

$y = a x + b x ^ { 2 }$
Where $a = \frac { v _ { 0 y } } { t v _ { x } }$ and $b = - \frac { v } { 2 \sigma _ { \log } ^ { 2 } }$ are constants.

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Video Transcript

you can say that the displacement in the extraction is gonna be equaling the initial velocity in the extraction multiplied by T we're going to then solve for tea. And so that t equals simply X divided by v x initial. So the exposition divided by V x initial. And so we can then say that the change in why or simply why is equaling v y initial t plus 1/2 g t squared. And so we can say that G in this case is gonna be negative so we can account for this and say negative 1/2 gt squared and we can then substitute in tea. So why is gonna be equaling the why initial multiplied by X divided by the ex initial minus 1/2 G multiplied by X over the ex initial quantity squared. And this is gonna be equaling V. Why initial over the ex initial Times X and then minus G. This would be G over, too. V X initial times x squared. And as you can see, this is in the form y equals a times X plus B x squared and then here a relating these two equations A is gonna be equaling V. Why initial over the ex initial, which is actually gonna be equaling two. The initial sign of fada over the initial co sign of data, which is equaling tangent of data Where needs to cancel out. So this would be equaling a, uh and then be is equaling negative g over to be ex initial squared. So this would be scared. My apologies. This would be squared because of this entire term is squared. So relating these two to our, um, standard form essentially y equals X plus B X squared. We confined the values for A and B, and this would be our answer for B. That is the end of the solution. Thank you for watching.

Carnegie Mellon University
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