Projectile Range on Level Ground

physicsmechanics/kinematicsdifficulty 2/5#projectile#kinematics#range#explorable

A projectile is launched from level ground at speed $v_0$ and angle $\theta$ above the
horizontal. Find its range $R$ (neglect air resistance).

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trajectory.

v0v0 θθ
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Answer

$ R = \frac{v_0^2 \sin(2\theta)}{g} $

Solution

The time of flight is $T = \tfrac{2 v_0 \sin\theta}{g}$, and the horizontal range is
$$R = v_0\cos\theta \cdot T = \frac{v_0^2 \sin 2\theta}{g},$$
which is largest at $\theta = 45^\circ$.