(adsbygoogle = window.adsbygoogle || []).push({}); 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. 3-46). (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 [Broken]

I need help on part b.

y_{0}= 1.5m

x = 20m.

Z = 10 degrees

2. Relevant equations

x = (v_{0}cos Z) t

0 = y_{0}+ (v_{0}sin Z)t - 0.5*g*t^{2}

3. The attempt at a solution

I have two equations in two variables, t and v_{0}. So I solved for v_{0}:

[tex]v_0^2 = (\frac{x^2}{2g\cos^2 Z})(\frac{1}{y_0 + x \tan Z})[/tex]

So, plugging in, I got 2.05 m/s for v_{0}. But that has to be wrong, especially since my answer in (a) is 36m/s.

What am I doing wrong?

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# Homework Help: Find the speed (2d)

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