Question 1083576
Neglecting friction, there is only one force acting on the bus as it moves up the ramp, 
and that is the gravitational force directed down the ramp, parallel to the incline.
This force is equal to m*g*sin(theta), where m is the mass of the bus.
The work done on the bus by this gravitational force is equal to the change in
kinetic energy of the bus in coming to a stop, W = delta-K
Since the final kinetic energy is zero, the change in kinetic energy is equal to the initial kinetic energy, K = (m/2)*v^2.
Let d = the distance traveled up the ramp
The work done is W = m*g*sin(theta)*d
W = K: m*g*sin(theta)*d = (m/2)*v^2
Solve for d: d = v^2/(2*g*sin(theta)) [Note that the distance is independent of mass]
Using MKS units, we first convert km/h to m/s: 72 km/h * 1000 m /km * 1 h/3600 s = 20 m/s
Putting in the numbers, we get d = 400/2/9.81/sin(8-deg) = 146.5 m