Question

which of these is the correct formula for the maximum range of a angular projectile?
a $v^{2}/g$
b $v^{2}/2g$
c $v/g$
d $2v/2g$

Explanation:

Step1: Recall range formula

The range formula for a projectile is $R=\frac{v^{2}\sin2\theta}{g}$, where $v$ is the initial velocity, $\theta$ is the angle of projection and $g$ is the acceleration due to gravity.

Step2: Find maximum - value condition

The function $y = \sin2\theta$ has a maximum value of 1. This occurs when $2\theta = 90^{\circ}$ or $\theta=45^{\circ}$.

Step3: Calculate maximum range

Substituting $\sin2\theta = 1$ into the range formula, we get $R_{max}=\frac{v^{2}}{g}$.

Answer:

A. $v^{2}/g$