
#1
Feb1110, 01:13 PM

P: 423

Hey I wondered what the best way to proceed about fiding the maximum lenght of a projectile when fired at an angle relative to an incline. Would a coordinate axis along the incline be the best choice? Then I could solve the equations for when y = 0, but it would also result in an acceleration in both dimentions.




#2
Feb1110, 02:18 PM

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Hi center o bass! Welcome to PF!
The hillside is linear, but your acceleration equations are quadratic, so keep the acceleration equations as simple as possible. 



#3
Feb1110, 02:36 PM

P: 423

Thank you! :) I'm from Norway and I was looking for some good norwegian physics forums, but I couldnt find any. This forum however seems very good with lots of activity and lots of diffrent topics.
Maximum height isn't really a problem when I think about it, but lenght is a bit tricky. So what if I assume that the projectile crosses the incline at the height y=h and i solve my second order equation for this time.. Then i subsitute this time into the component desciribing the xposition giving me x as a function of the angle? What do you think? 



#4
Feb1110, 02:56 PM

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Maximum height and length on an incline.What do you get? 



#5
Feb1110, 03:23 PM

P: 423

The ycomponent of the position is give by
[tex]y(t) = v_0 \sin \theta t  \frac{1}{2} g t^2 = h[/tex] solving this for t i get [tex] t* = \frac{v_0 \sin \theta \pm \sqrt{v_0^2 \sin^2 \theta 2gh}}{g}[/tex] substituting this into [tex]x(t) = v_0 \cos \theta t[/tex] and assuming its the positive root that is valid i get [tex]x(t*) = \frac{1}{g} (v_0 \cos \theta \sin \theta + \cos \theta \sqrt{v_0^2 \sin^2 \theta  2gh}) = \frac{1}{g} \left( \frac{1}{2} v_0 \sin 2\theta + \sqrt{\frac{1}{4} v_0^2 \sin^2 2\theta  2gh}\right)[/tex] giving that the angle that maxemises the expression is [tex]\frac{\pi}{2}[/tex] which is reasonable and the same as for a projectile shot out from a straight line. Is this correct? :) 



#6
Feb1110, 03:36 PM

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Sorry, I'm confused … where have you used y/x = slope of the hill?
(and π/2 is vertical) 



#7
Feb1110, 03:56 PM

P: 423

Sorry. It is ofcourse [tex]\frac{\pi}{4}[/tex] that maxemises the expression. I have not used the slope I assumed that the angle was bigger than the angle of the hill so that the projectile would follow a path and hit the hill at a height h.
In simplifying the expression I have used the trigonometric identity [tex]\sin \theta \cos \theta = \frac{1}{2}\sin 2\theta[/tex]. Are you with me? How would one use the slope of the hill? 



#8
Feb1210, 03:45 PM

P: 423

I am unsure of this, so it would be great if anyone confirmed if this is true or not :)



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