**1. The problem statement, all variables and given/known data**

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3–41). (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 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 7.0° above the horizontal, what is the new minimum speed?

**2. Relevant equations**

[tex]y=y_0+v_0t+.5at^2[/tex]

[tex]v^2=v_0^2+2a(y-y_0)[/tex]

[tex]x=v_0t[/tex]

**3. The attempt at a solution**

The answer to part a is obviously 40m/s, but the answer to b eludes me.

I need v, so I defined v_x in terms of v via the kinematic equation [tex]x=v_0t[/tex]

and then solved it for t

[tex]22=tcos\theta[/tex]

[tex]t=\frac{22}{vcos\theta}[/tex]

However, when I attempt to do this with v_y, I end up with a nasty quadratic

[tex]-4.9t^2+tvsin\theta+1.5=0[/tex]

and solving this for t does not seem to be advisable.

http://ift.tt/1fAtoCz