## College Physics (4th Edition)

The correct answer is: (b) $6h$
We can assume that the kangaroo's initial vertical velocity $v_{0y}$ is the same on the Earth and on the moon. We can find an expression for the kangaroo's height on Earth: $v_{yf}^2 = v_{0y}^2+2a_yh$ $h = \frac{v_{yf}^2 - v_{0y}^2}{2a_y}$ $h = \frac{0 - v_{0y}^2}{(2)(-g)}$ $h = \frac{v_{0y}^2}{2g}$ We can find an expression for the kangaroo's height on the moon: $v_{yf}^2 = v_{0y}^2+2a_yh_m$ $h_m = \frac{v_{yf}^2 - v_{0y}^2}{2a_y}$ $h_m = \frac{0 - v_{0y}^2}{(2)(-g/6)}$ $h_m = 6\times\frac{v_{0y}^2}{2g}$ $h_m = 6h$ The correct answer is: (b) $6h$