physics
posted by perry .
A stunt driver wants to make his car jump over eight cars parked side by side below a horizontal ramp.
(a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5m above the cars, and the horizontal distance he must clear is 20m.
(b) If the ramp is now tilted upward, so that "takeoff angle" is 10 degrees above the horizontal, what is the new minimum speed?

this is from that white book that i'm also using...
the odd answers in the book are in the back in which this problem is one I belive 
You know the horizontal distance it must go.
Horizontal distance=vi*time
verticalpositionfinal=verticalpositioninitial 4.9 t^2
0=1.54.9 t^2 solve for t, then put that into the horizontal distance equation and solve for vi.
Now if you tilt it, then the vihoriontal is viCos10, and vivertical is not zero, but ViSin10.
Work it the same way, it is a little tricker on the algebra, but it works cleanly. 
I don't know how the author of the Giancoli text reached 20 m/s as the answer to the second part of this questions. Nevertheless, I do know that the solution Bobpursley suggests here doesn't work. I find it interesting that there isn't one straightforward solution to this question on the entire internet. I've been working on the problem for two days, and none of the strategies I have tried give me 20 m/s.

To call the algebra on this question convuluted is a profound understatement. Nevertheless, I have solved it.
The range equals 20 meters and the initial velocity of the car along the xaxis is 0.98Vi (or the Vicos10). Since the combined time the car travels up and down on the yaxis equals the time the car travels the entire way along the xaxis, you can set the equation as follows:
time = time
or
20 / 0.98Vi = time on yaxis
Divide 20 by 0.98 to find that the ratio of the initial velocity is:
20.4 / Vi
...or more simply put, the initial velocity if 20 m/s.