Answer
(a) $v_{\circ}=\sqrt{gr+2gh}$
(b) The car will leave the surface of the road.
Work Step by Step
(a) We can find the required initial speed as follows:
We know that the net force on the car is given as
$\Sigma F=\frac{mv^2}{r}$
$mg=\frac{mv^2}{r}$
$\implies v^2=gr$
We plug in this value in the equation of motion:
$v_i^2-v_f^2=2gh$
$\implies v_{\circ}^2-gr=2gh$
This simplifies to:
$v_{\circ}=\sqrt{gr+2gh}$
(b) We know that if the value of the initial speed of the car exceeds the value determined in part (a), then the car will leave the surface of the road.
