Mechanics: calculating launch angle of projectile

[PeroK]

You can invent your own equation in terms of the given quantities. Knowing the time of flight makes the task easy. You know that the maximum height occurs at half the time of flight and that the vertical velocity is zero when that happens. Then you can write two equations expressing these ideas:
$$h_{\text{max}}=v_{\text{0y}}\frac{t_{\!f}}{2}-\frac{1}{2}g\left(\frac{t_{\!f}}{2}\right)^2~;~~0=v_{\text{0y}}^2-2gh_{\text{max}}.$$The task before you is to use the second equation to substitute $v_{\text{0y}}$ into the first equation and solve the ensuing quadratic.

You do need to be careful if $t_f$ is not when it returns to the ground.

 

[kuruman]

You do need to be careful if $t_f$ is not when it returns to the ground.

Oops! I lost myself somewhere.