Projectile motion on an inclined plane

[Maianbarian]

Homework Statement

A projectile is fired from the origin down an inclined plane that makes an angle theta with the horizontal. The projectile is launched at an angle alpha to the horizontal with an initial velocity v.

Show that the angle of elevation alpha that will maximise the the downhill range is the angle halfway between the plane and the vertical.

Homework Equations

By taking the x-axis to be down the plane and the y-axis to be perpendicular to the plane I managed to get an expression for the range along the plane to be:

R= 2v^2(sin(\alpha+\theta)/(gcos^2(\theta))

The Attempt at a Solution

I differentiated the above equation w.r.t. alpha and set to zero to get the max. solving for alpha I got

\alpha= \pi/(2) -\theta

I don't see how this is halfway between the plane and the vertical, if someone could explain this to me I would be most grateful, or if my value for alpha is wrong could someone please point me in the right direction? :)

 

[vela]

Your expression for the range is incorrect. Did you remember that $a_x \ne 0$ when you rotate the axes? You should get
$$R = \frac{2v^2}{g \cos\theta} \sin(\alpha+\theta)\cos\alpha.$$ Maximizing $R$ as a function of $\alpha$ will lead to the desired result.