Question

A 1998 Ford Mustang (1460. kg)  is coasting up a hill moving at 49.1 m/s, 10. m above the bottom of the hill. How fast will the car be moving when it reaches the top of the hill 12 m above the bottom of the hill?

Answer 1

So, the car will go 2 m up. According to the conservation of energy, some part of its kinetic energy will be converted into work required to increase the potential energy by 2 m:

$$\frac12 mv^2=\frac12mu^2+mgh,\\\space\\ u=\sqrt{v^2-2gh}=\sqrt{49.1^2-2·9.8·2}=48.7\text{ m/s}.$$