The vertical section of a road over a flyover bridge in the direction of its length is in the form of an arc of a circle of radius $ {\text{19}}{\text{.5m}} $ . Find the maximum speed at which a truck can cross the bridge without losing contact with the road at its highest point, if the centre of gravity of the truck is $ 0.5m $ above the road surface. ( $ g = 9.8{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ ).

Hint : Two forces, one of weight and the other centripetal force, acts on the truck. By applying the concepts of circular motion to solve the question and then by equating centripetal force to weight we will get the answer.

Formula Used: We will be using the following formula,
Centripetal force $ {F_c} = \dfrac{{m{v^2}}}{r} $ where $ m $ is the mass, $ v $ is the velocity and $ r $ is the radius of circle.
Weight $ w = mg $ where $ m $ is the mass and $ g $ is the acceleration due to gravity.

Complete step by step answer
 It has been given that the vertical section of a road over a flyover bridge in the direction of its length is in the form of an arc of a circle of radius $ {\text{19}}{\text{.5m}} $ . We need to find the maximum speed at which a truck can cross the bridge without losing contact with the road at its highest point, if the centre of gravity of the truck is $ 0.5m $ above the road surface.
Centripetal force acts on the truck in a horizontal direction. A centripetal force is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous centre of curvature of the path. Mathematically, centripetal force $ {F_c} = \dfrac{{m{v^2}}}{r} $ where $ m $ is the mass, $ v $ is the velocity and $ r $ is the radius of circle.
The weight of the truck, which is equivalent to the product of mass of the truck and the acceleration due to gravity, acts vertically downwards, perpendicular to the body.
Mathematically, weight $ w = mg $ where $ m $ is the mass and $ g $ is the acceleration due to gravity. Given, $ g = 9.8{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ .
From equation of vertical circular motion, we may write,
 $ mg = \dfrac{{m{v^2}}}{r} $
 $ \Rightarrow v = \sqrt {rg} $
It has been given, radius $ {\text{r = 19}}{\text{.5m}} $ and $ g = 9.8{m \mathord{\left/
 {\vphantom {m {{s^2}}}} \right.} {{s^2}}} $ .
Thus, $ v = \sqrt {19.5 \times 9.8} = 13.9{m \mathord{\left/
 {\vphantom {m {s.}}} \right.} {s.}} $ .

Note
There is another force that is acting on the truck. The normal reaction force acts perpendicular to the body, in a direction, exactly opposite to the weight. In other words, it is a force that is directed perpendicular to the two surfaces in contact.
Normal force is a contact force. If two surfaces are not in contact, they can't exert a normal force on each other. The normal force is the force that surfaces exert to prevent solid objects from passing through each other.