# Trigonometric Optimization

I am trying to solve a this problem: An object of mass $m$ is launched at angle $\theta$ from the horizontal with initial velocity $\underline{u}$. The ground however is on a slope of $\alpha$ with the horizontal.

I've used some reasoning and integration, finding (then I also checked it is correct on the book) that the trajectory is: $$y(x) = x\tan{\theta}-\frac{gx^2\sec^2{(\theta)}}{2U^2}$$ Now I want to find the function describing when the object will touch the slope and then maximise it.

So my idea was to write the slope as $y(x) = x\tan{\alpha}$ and then substitute into the above and find: $$x = \frac{2U^2\cos^2{\theta}(\tan{\theta}-\tan{\alpha})}{g}$$

However I have no idea on how to optimize this function. I tried using Mathematica with the Maximize function, but it didn't work.

I thought that I should check when $\tan{\theta}-\tan{\alpha}$ is positive, however it doesn't solve the problem. I tried using some trigonometry and I could rewrite the function as: $$x = \frac{2U^2\cos{(\theta)}\sin{(\theta-\alpha)}}{g\cos{(\alpha)}}$$ But nothing new came out of this. Could you suggest any other procedure?

I want to optimize the range calculated on the slope! So at which angle $\theta$ I should shoot the object to get the furthest away.

I've tried to use a suggestion and instead of using the normal axis, I am using the coordinate system parallel to the slope. So my idea would be to take the equation of the range in the case without slope (i.e. $x = \frac{2U^2\sin{(2\beta)}}{g}$ where $\beta$ would be the angle with the horizontal) and then simply modify it for the new coordinate system. So that $\beta = \theta - \alpha$ . However in this way I get $x = \frac{2U^2\sin{(2(\theta-\alpha))}}{g}$.

To optimize it we need to have $\sin{2(\theta-\alpha)} = 1$, which implies $\theta = \frac{\pi}{4} + \alpha$.

This is very close to the solution which is $\theta = \frac{\pi}{4} + \frac{\alpha}{2}$. Where have I made a mistake?

Thank you!

• Find the roots of the derivative of $\cos^2\theta(\tan\theta-\tan\alpha)$. – Yves Daoust Oct 31 '16 at 21:42
• But you forgot to tell us what you want to optimize. – Yves Daoust Oct 31 '16 at 21:43
• Are you trying to find the value of $\theta$ for which the distance up the slope is maximized? – David Quinn Oct 31 '16 at 21:44
• I've edited it now, I want to optimize the range on the slope! – Euler_Salter Oct 31 '16 at 21:46
• It's easier to solve if you set up the equations of motion with the axes aparallel to and perpendicular to the slope. Check out Projectile Motion on an Inclined Plane. – David Quinn Oct 31 '16 at 21:49