An object is projected in the air with initial velocity u at an...

The maximum range is given by, {R}_\max = \dfrac{{u}^{2}\sin 2\left(45 °\right)}{g} = \dfrac{{u}^{2}}{g}.

For the range to be half of the maximum range,

\dfrac{{R}_\max}{2} = \dfrac{{u}^{2}}{2g} = \dfrac{{u}^{2}\sin 2\theta}{g}

$⇒ \sin 2\theta = \dfrac{1}{2}$

$⇒ 2\theta = 30 ° \text{or}2\theta = 150 °$

$⇒ \theta = 15 ° , 75 °$

Therefore, answer is $15\text{or}75$