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**1. 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 [Broken]

I need help on part b.

y

_{0}= 1.5m

x = 20m.

Z = 10 degrees

**2. Homework 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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