Answer:
The correct answer is option b
Explanation:
The range of a projectile is given by,
[tex]R=\frac{v^{2}sin 20 }{g} [/tex]
Where
v is the initial velocity with which the projectile was launched.
θ is the angle of the projection projectile.
g is the acceleration due to gravity.
The range will be maximum when the term sin2θ assumes the maximum possible value.
The maximum possible value of sine of an angle is 1. Thus the maximum range is given by,
[tex]R=\frac{v^{2}x 1 }{g}\\
=\frac{v^{2} }{g} [/tex]
Therefore the correct answer is option b, v²/g
