Hi there!
Ignoring friction, we know that the centripetal force experienced by the car is due to the normal force exerted by the road.
We can do a summation of forces in both the horizontal and vertical directions.
Vertical:
[tex]W = Mg[/tex], force due to gravity
[tex]Ncos\theta[/tex], VERTICAL component of the normal force.
[tex]\Sigma F_y = Ncos\theta - Mg\\\\Mg = Ncos\theta[/tex]
Horizontal:
[tex]Nsin\theta = F_{Hnet}[/tex]
The net horizontal force is equivalent to the centripetal force:
[tex]Nsin\theta = \frac{mv^2}{r}[/tex]
We can solve for theta by dividing:
[tex]\frac{Nsin\theta = \frac{mv^2}{r}}{Ncos\theta = mg}[/tex]
Simplify:
[tex]tan\theta = \frac{ \frac{v^2}{r}}{ g}\\\\tan\theta =\frac{v^2}{rg}[/tex]
Solve:
[tex]\theta = tan^{-1}(\frac{v^2}{rg}) = tan^{-1}(\frac{30^2}{(1000)(9.8)}) = \boxed{5.26^o}[/tex]
