For a projectile launched with speed $v_0$ at angle $\theta$, the horizontal range is
A$R=\dfrac{2v_0\sin\theta}{g}$
B$R=\dfrac{v_0^2\sin2\theta}{g}$
C$R=\dfrac{v_0^2\sin^2\theta}{2g}$
D$R=\dfrac{v_0\cos\theta}{g}$
Answer & Solution
Correct answer: B. $R=\dfrac{v_0^2\sin2\theta}{g}$
The standard expression for the horizontal range of a projectile on level ground is $R=\dfrac{v_0^2\sin2\theta}{g}$. Option A is the time of flight, and option C is related to maximum height.
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