You probably know that (for reasonable $t$) the vertical distance travelled by the first object after $t$ seconds is $\frac{1}{2}at^2$, where $a$ is the acceleration. So if $t_1$ is the time until the first object hits the ground, we have $$26.83=\frac{1}{2}(9.81)t_1^2.$$ Now we can compute $t_1$.
Look at the second object. Since time of flight and horizontal distance travelled is the same as for the first object, the horizontal speed $v_1$ of the second object is $23.44$.
Let the initial vertical speed be $v_2$. Then after time $t_1/2$, the second object reached maximum height. The vertical velocity of the second object, after time $t$, where $t\le t_1$, is equal to $v_2-9.81t$. At time $t_1/2$, the vertical speed reaches $0$, and therefore $$v_2-9.81t_1/2=0.$$ Since we know $t_1$, we now know $v_2$.
The launch angle $\theta$ is easy to compute once we know $v_1$ and $v_2$, because $$\tan\theta=\frac{v_2}{v_1}.$$