Answer:
[tex]v=28m/s[/tex]
Explanation:
The vertical component of the normal force must cancel out with the weight of the car taking the curve:
[tex]N_y=W[/tex]
[tex]Ncos\theta=mg[/tex]
(Notice it has to be cos and not sin, because the angle [tex]\theta[/tex] is the slope, for null slope [tex]N_y=Ncos(0)=N[/tex], as it should be).
The horizontal component of the normal force must be the centripetal force, that is:
[tex]N_x=F_{cp}[/tex]
[tex]Nsen\theta=ma_{cp}[/tex]
[tex](\frac{mg}{cos\theta})sin\theta=m\frac{v^2}{r}[/tex]
[tex]gtan\theta=\frac{v^2}{r}[/tex]
[tex]v=\sqrt{grtan\theta}=\sqrt{(9.8m/s^2)(80m)tan(45^{\circ})}=28m/s[/tex]
