# Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 46

The two possible solutions are $\theta = 15^{\circ}$ and $\theta = 75^{\circ}$.

#### Work Step by Step

$range = \frac{v_0^2~sin(2\theta)}{g}$ The maximum possible range occurs when $sin(2\theta) = 1$, that is, when $\theta = 45^{\circ}$. In that case, $range = \frac{v_0^2}{g}$. We need to find $\theta$ such that $range = \frac{1}{2}\times \frac{v_0^2}{g}$ $\frac{v_0^2~sin(2\theta)}{g} = \frac{1}{2}\times \frac{v_0^2}{g}$. Therefore; $sin(2\theta) = 0.5$ $2\theta = arcsin(0.5)$ $2\theta = 30^{\circ}$ $\theta = 15^{\circ}$ Note that: $sin(180^{\circ}-2\theta) = sin(2\theta)$ We can find another possible solution. $180^{\circ}-2\theta = arcsin(0.5)$ $180^{\circ}-2\theta = 30^{\circ}$ $\theta = 75^{\circ}$ The two possible solutions are $\theta = 15^{\circ}$ and $\theta = 75^{\circ}$.

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