#### Answer

$\theta=45^{\circ}$

#### Work Step by Step

Let angle of initial velocity be $\theta$ with respect to the horizontal.
$v^{2}=u^{2}+2aH$
$0=(usin\theta)^{2}-2gH$
$H=\frac{u^{2}sin\theta^{2}}{2g}$
Now, $v=u+at$
$0=usin\theta-gt$
$t=\frac{usin\theta}{g}$
So, $R=ucos\theta \times t=ucos\theta \times \frac{usin\theta}{g}$
Therefore, $H=\frac{1}{2} R$
or, $\frac{u^{2}sin\theta^{2}}{2g}=\frac{1}{2}ucos\theta \times \frac{usin\theta}{g}$
or, $tan\theta=1$
So, $\theta=45^{\circ}$