Tough calc projectile Q w/ incline. HELP

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Homework Help Overview

The problem involves a projectile motion scenario where a ball is launched at an angle \(\theta\) from the base of an inclined hill at an angle \(\phi\). The objective is to determine the angle \(\theta\) that maximizes the horizontal distance \(x\) at which the ball strikes the hill.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between vertical displacement and the angle of launch, with attempts to express the equations of motion in terms of \(x\) and \(\theta\). There are questions about how to proceed after deriving certain equations and how to handle derivatives in the context of maximizing the distance.

Discussion Status

The discussion is ongoing, with participants offering various approaches to derive the function \(x(\theta)\) and suggesting the use of calculus to find its maximum. There is an acknowledgment of potential errors in earlier equations, prompting further exploration of the relationships between the variables.

Contextual Notes

Participants are working under the constraints of projectile motion equations and the specific geometry of the problem, including the angles of launch and incline. There is a focus on ensuring that all variables are treated correctly when taking derivatives.

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A ball is hit at an angle [tex]\theta[/tex] to the horizontal, from the base of a hill [tex]\phi[/tex] degrees in incline. The ball strikes the hill at a horizontal distance [tex]x[/tex] measured from the launch point. Determine the value of [tex]\theta[/tex] that will result in the largest possible value of [tex]x[/tex].

well i started out finding the vertical displacement to be [tex]x\tan\phi[/tex]
then i put that in the basic [tex]S_{y} = V_{i}sin\theta t + \frac{1}{2}-9.8t^2[/tex] of corse [tex]S_{y} = xtan\phi[/tex]. now i am stumped on what to do. please help!
 
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Let y_1 = x tan ph and
Let y_2 = v_0(sin th)t - (1/2)gt^2, x = v_0(cos th)t

note that t can be eliminated from y_2 by putting t = x/[ v_0 cos th ]
 
so then i come up with [tex]y_2 = x tan \theta - \frac {4.9x^2}{(v_0^2)(cos /theta^2)}[/tex] if i set that equal to [tex]x tan \phi[/tex] , i can simplify to [tex]tan \phi = tan \theta - \frac {4.9x}{(v_0^2)(cos^2 \theta)}[/tex]

what would i do next??
 
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right, solve the equation you just found for x. Note that this is really a function x(th) that gives you the horz distance x at which the ball hits the hill when thrown at an angle th. Now you can use calculus methods to find the max of this function. be very careful when using inverse functions (like arctan). you might find the following relation very helpful as well, 2 sin th cos th = sin 2 th, use it when you can
 
ok i got : x = (v_0/g) (v_0 sin2th - 2y cos th). if i take the derivative of his equation (velocity) how would i find the derivitve of "y" ??
 
i'm not sure the equation you got is correct (it might be..), but if you look at your earlier post, you will see an equation with an 'x' but no 'y'. Just solve this for 'x' and you have the function x(th).

btw, because you want to maximize x(th) with respect to theta, make sure to take the derivative with respect to th and notice that all the other variables remain constant in the problem as th changes, so you can treat them as constants in computing the derivative.
 
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