Answer:
[tex]\theta = 15^o \: or\: 75^o[/tex]
Explanation:
As we know that the formula of range is given as
[tex]R = \frac{v^2sin2\theta}{g}[/tex]
now we know that
maximum value of the range of the projectile is given as
[tex]R_{max} = \frac{v^2}{g}[/tex]
now we need to find such angles for which the range is half the maximum value
so we will have
[tex]\frac{R}{2} = \frac{v^2}{2g} = \frac{v^2sin(2\theta)}{g}[/tex]
[tex]sin(2\theta) = \frac{1}{2}[/tex]
[tex]2\theta = 30 or 150[/tex]
[tex]\theta = 15^o \: or\: 75^o[/tex]
