# Maximum height and length on an incline.

by center o bass
Tags: projectile
 P: 471 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.
HW Helper
Thanks
P: 26,148
Hi center o bass! Welcome to PF!
 Quote by center o bass 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.
 P: 471 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?
HW Helper
Thanks
P: 26,148
Maximum height and length on an incline.

 Quote by center o bass 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?
 P: 471 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? :)
 Sci Advisor HW Helper Thanks P: 26,148 Sorry, I'm confused … where have you used y/x = slope of the hill? (and π/2 is vertical)
 P: 471 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?
 P: 471 I am unsure of this, so it would be great if anyone confirmed if this is true or not :)