Answer
60 m/s at 61 degrees.
Work Step by Step
Let the origin be the launch point from which the projectile is launched, and let upward be the positive y direction.
Find the horizontal velocity component.
$$v_x=\frac{195\;m}{6.6\;s}=29.55m/s$$
Now find the initial y velocity.
$$y=y_0+v_{y0}t+0.5a_yt^2$$
$$135m= v_{y0}(6.60s)+0.5(-9.8m/s^2)(6.6s)^2$$
$$v_{y0}=52.79m/s$$
Now that we have the components, find the speed of 60 m/s by using the Pythagorean Theorem.
$$v=\sqrt{{v_{0y}^2+}{v_{0x}^2}}\approx 60\;m/s$$
Then find the angle.
$$\theta = \tan^{-1}\frac{v_{0y}}{v_{0x}}=61^{\circ}$$
