Maximum height and length on an incline.


by center o bass
Tags: projectile
center o bass
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#1
Feb11-10, 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.
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tiny-tim
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Feb11-10, 02:18 PM
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Hi center o bass! Welcome to PF!
Quote Quote by center o bass View Post
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?
Nah too complicated.

The hillside is linear, but your acceleration equations are quadratic, so keep the acceleration equations as simple as possible.
center o bass
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#3
Feb11-10, 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 x-position giving me x as a function of the angle? What do you think?

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Feb11-10, 02:56 PM
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Maximum height and length on an incline.


Quote Quote by center o bass View Post
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 x-position giving me x as a function of the angle? What do you think?
Yes, doing y first, finding t, then doing x (or the other way round), should be fine.
What do you get?
center o bass
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#5
Feb11-10, 03:23 PM
P: 423
The y-component 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? :)
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Feb11-10, 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)
center o bass
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#7
Feb11-10, 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?
center o bass
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#8
Feb12-10, 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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Feb12-10, 03:58 PM
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Quote Quote by center o bass View Post
How would one use the slope of the hill?
If the slope of the hill is k, so that y = kx, then you need to get two equations, one for x and t, and one for y and t, and then eliminate t and put y = kx.


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