A body is released from rest at the top of a frictionless inclined plane of height 5 m. Taking g = 10 m/s², what is the speed of the body at the bottom of the incline?

By conservation of energy, all potential energy converts to kinetic energy at the bottom: mgh = ½mv², so v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. Option A results from using v = √(gh) (forgetting the factor of 2), option C from using v = 2gh/2 without the square root, and option D from using v = 2gh.