Minimum speed for ramp jump at 10 degree takeoff angle

[endeavor]

Homework Statement

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3-46). (a) With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 m above the cars and the horizontal distance he must clear is 20m. (b) What is the new minimum speed if the ramp is now titled upward, so that "takeoff angle" is 10 degrees above the horizontal, and nothing else is changed?

http://img181.imageshack.us/img181/6396/chp3pro46sk5.th.png

I need help on part b.
y₀ = 1.5m
x = 20m.
Z = 10 degrees

Homework Equations

x = (v₀ cos Z) t
0 = y₀ + (v₀ sin Z)t - 0.5*g*t²

The Attempt at a Solution

I have two equations in two variables, t and v₀. So I solved for v₀:
$$v_0^2 = (\frac{x^2}{2g\cos^2 Z})(\frac{1}{y_0 + x \tan Z})$$

So, plugging in, I got 2.05 m/s for v₀. But that has to be wrong, especially since my answer in (a) is 36m/s.

What am I doing wrong?

 

[endeavor]

I know what I did wrong. I put g in the denominator, but it should be in the numerator...

 

[chanvincent]

algebra mistake... the g should be placed on the numerater...

 

[drewtoid]

How did you develop the equation to solve for the initial velocity?