Chapter 3 - Exercises and Problems - Page 48: 39

Work Step by Step

We know that $R=\frac{v_{\circ}^2}{g}$ Hence, the range is inversely proportional to g $\frac{R_{Moon}}{R_{Earth}}=\frac{g_{Earth}}{g_{Moon}}$ $R_{Moon}=R_{Earth}(\frac{g_{Earth}}{g_{Moon}})$ We plug in the known values to obtain: $R_{Moon}=(180)(\frac{9.8}{1.63})=1100m$

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