Question
A projectile's horizontal range on level ground is $R=$ $v_{0}^{2} \sin 2 \theta / g .$ At what launch angle or angles will the projectile land at half of its maximum possible range.
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The maximum range is achieved when the launch angle $\theta$ is $45^{\circ}$. Show more…
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The range $R$ of a projectile fired with an initial velocity $v_{0}$ at an angle $\theta$ with the horizontal is $R=\frac{v_{0}^{2} \sin 2 \theta}{g}$, where $g$ is the acceleration due to gravity. Find the angle $\theta$ such that the range is a maximum.
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The range of a projectile is $R=\frac{v_{0}^{2} \sin 2 \theta}{g},$ where $v_{0}$ is its initial velocity, $g$ is the acceleration due to gravity and is a constant, and $\theta$ is its firing angle. Find the angle that maximizes the projectile's range.
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