Accelerating onto and over inclined plane Consider an RC car going off a jump.  What angle is optimal to achieve the greatest distance?  The height of each jump is the same.  The angle of the jump and the related length of the jump changes.  This is an experiment my son is doing for his 6th grade science fair.
The RC car used is a hobbyist RC car with a top speed of approximately 30 mph.  It has shocks, and the tires are rubber with foam inside.
We've used a Netduino to control acceleration so that the each run is identical.


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*The optimal angle for ballistic trajectory is 45 degrees.  This is based on his research as he hasn't quite reached calculus yet.  http://en.wikipedia.org/wiki/Range_of_a_projectile

*My understanding is that the Work to climb the inclined is identical for each jump.  They are all the same height, and climbing them produces the same potential energy.  Am I completely misunderstanding this?

*The tires do not slip.

*The car starts from 20 feet from the high end of the jump.
Given this information, would the 45 degrees be the optimal angle of the jump?  We haven't completed the test runs yet, but initial tests show the 45 degree jump performs poorly.
We reduced the test speed such that the car does not scrape the bottom when it hits the ramp.
Other than the trajectory and the Work required to climb the jump, what aspects are we missing?  Where is the energy going when it hits the 45 degree jump?
 A: If you are limiting the takeoff speed to prevent it bottoming out, then I suggest you lower the ramp. $45^\circ$ gives optimal range for a given takeoff speed (ignoring friction), but only if you don't care what the vertical component of the velocity is on landing.
At $30\: \mathrm{mph}$, with a $45^\circ$ jump, you say it doesn't bottom out. The vertical component of the velocity on landing has approximately the same magnitude as on take off (air resistance losses), which would be $30 \; \cos(45^\circ)$ = about $26\: \mathrm{mph}$.
To maximise the range, then, we need to keep this vertical component ($v_y = 26\: \mathrm{mph}$), whilst letting the overall speed ($v$) increase to $45\: \mathrm{mph}$. 
$v^2 = v_x^2 + v_y^2$
$v_x^2 = v^2 - v_y^2 = 45^2 - 26^2 = 1349$
$v_x = \sqrt{1349} = 36.7 \mathrm{mph} $
So we can calculate the angle from the $x$ and $y$ components of the velocity:
$\tan(\theta) = v_y / v_x = 26 / 36.7 = 0.71$
$\theta = \arctan(\theta) = \arctan(0.71) = 35^{\circ}$
So based on that information, try $35^\circ$, with a speed at the ramp of $45\: \mathrm{mph}$.
