Answer to Question #102180 in Mechanics | Relativity for sdfsd

Question

A 67.9 kg skier is at rest at the top of a 298 m high run. The slope is inclined at a 15.7º angle relative to horizontal. Assuming that there is no friction between the skis and the snow, how fast is the skier traveling when he/she is halfway down the slope?

Answer 1

The total mechanical energy at the top of the slope is equal

$$E=mgh$$ $$=67.9\:\rm kg\times 9.8\: N/kg\times 298\: m=198\:000\;J.$$

Since there is no friction between the skis and the snow, the total energy is conserved. Hence, at the halfway down the slope the skier have the same amount of energy as at the beginning of motion.

$$E=mgh/2+mv^2/2=E/2+mv^2/2.$$

So

$$mv^2/2=E/2=198\:000/2=99\:000\:\rm J.$$

$$v=\sqrt{2\times 99\:000/67.9}=54\:\rm m/s.$$