## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 6 - Problems - Page 229: 44

#### Answer

The meteor's speed when it hits the surface of the planet is 13.0 km/s

#### Work Step by Step

If the meteor's speed far away was zero, the kinetic energy it would gain before striking the planet's surface would be $\frac{1}{2}mv_e^2$. When the meteor's speed far away is non-zero, it will still gain the same amount of kinetic energy before it strikes the planet's surface. We can find the meteor's speed when it hits the surface of the planet: $KE = \frac{1}{2}mv_0^2+\frac{1}{2}mv_e^2$ $\frac{1}{2}mv^2 = \frac{1}{2}mv_0^2+\frac{1}{2}mv_e^2$ $v^2 = v_0^2+v_e^2$ $v = \sqrt{v_0^2+v_e^2}$ $v = \sqrt{(5.0~km/s)^2+(12.0~km/s)^2}$ $v = 13.0~km/s$ The meteor's speed when it hits the surface of the planet is 13.0 km/s.

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