An astronaut drops a rock from the top of a crater on the Moon. When the rock is halfway down to the bottom of the crater, its speed is what fraction of its final impact speed?
A) $\displaystyle \frac{1}{4}$
B) $\displaystyle \frac{1}{2\sqrt{2}}$
C) $\displaystyle \frac{1}{2}$
D) $\displaystyle \frac{1}{\sqrt{2}}$
A) $\displaystyle \frac{1}{4}$
B) $\displaystyle \frac{1}{2\sqrt{2}}$
C) $\displaystyle \frac{1}{2}$
D) $\displaystyle \frac{1}{\sqrt{2}}$
Because the rock has lost half of its gravitational potential energy, its kinetic energy at the halfway point is half of its kinetic energy at impact. Since $K$ is proportional to $v^2$, if $K_{\mathrm{at\ halfway\ point}}$ is equal to $K_{\mathrm{at\ impact}}$, then the rock’s speed at the halfway point is $\sqrt{\displaystyle \frac{1}{2}}=\displaystyle \frac{1}{\sqrt{2}}$ its speed at impact.