Conservation of Energy problem...help?

Question

A runaway truck lane heads uphill at 30° to the horizontal. If a 16000 kg truck goes out of control and enters the lane going 115 km/h, how far along the ramp does it go? Neglect friction.

Answer 1

Truck's KE = mv^2/2. When it has stoped, KE is converted to PE = mgh. So mv^2/2 = mgh. Both sides can be divided by m (thus note m is irrelevant), and we have v^2 = 2gh, h = v^2/(2g). For distance along ramp, divide h by sin(theta).