Maximum height and length on an incline.

[center o bass]

Hey I wondered what the best way to proceed about fiding the maximum length 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.

 

[tiny-tim]

Welcome to PF!

Hi center o bass! Welcome to PF!

Hey I wondered what the best way to proceed about fiding the maximum length 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]

Thank you! :) I'm from Norway and I was looking for some good norwegian physics forums, but I couldn't 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 length 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?

 

[tiny-tim]

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]

The y-component of the position is give by

y(t) = v_0 \sin \theta t - \frac{1}{2} g t^2 = h

solving this for t i get

t* = \frac{v_0 \sin \theta \pm \sqrt{v_0^2 \sin^2 \theta -2gh}}{g}

substituting this into x(t) = v_0 \cos \theta t and assuming its the positive root that is valid i get

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)

giving that the angle that maxemises the expression is \frac{\pi}{2} which is reasonable and the same as for a projectile shot out from a straight line. Is this correct? :)

 

[tiny-tim]

Sorry, I'm confused … where have you used y/x = slope of the hill?

(and π/2 is vertical)

 

[center o bass]

Sorry. It is ofcourse \frac{\pi}{4} 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 \sin \theta \cos \theta = \frac{1}{2}\sin 2\theta.

Are you with me?
How would one use the slope of the hill?

 

[center o bass]

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

 

[tiny-tim]

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.