At what angle must a projectile be fired to travel 3.624m from a height of 1.012m?

x=3.624m y=1.012m v0=? t=? v=?

$H = \frac{{U}^{2} {\sin}^{2} \theta}{2} g$ , $R = \frac{{U}^{2} \sin 2 \theta}{g}$ , $t = \frac{2 U \sin \theta}{g}$
Here $H$ is the height of the projectile along $Y$ axis and $R$ is the range along $X$ axis. $U$ is the initial velocity and $\theta$ is the angle of projectile thrown.
Here we get , $H = 1.012 m$ and $R = 3.624 m$