G1470-kilogram truck moving with a speed of 24.0 m/s runs into the rear end of a 1110-kilogram stationary car. if the collision is completely inelastic, how much kinetic energy is lost in the collision?

Kinetic energy = Potential energy 

(1/2)mV^2 = mgh 

where 

m = mass of the car 
V = velocity of the car = 15 m/s (given) 
g = acceleration due to gravity = 9.8 m/sec^2 (constant) 
h = vertical distance travelled by car = d(sin 19) 
d = distance travelled by the car along the incline 

Substituting values, 

(1/2)(m)(529) = m(9.8)h 

and since "m" appears on both sides of the equation, it will simply cancel out, hence the above becomes 

(1/2)(529) = 9.8h 

h = 225/19.6 = 26,98 m 

Let 

d = distance that the car will coast up the hill before coming down 

h/d = sin 19 

and therefore, 

d = h/sin 19 

and substituting values, 

d = 26,98/sin 19 

d = 82,870 m --- after coasting up the hill for this distance, the car will start to roll down.